Class 8th Mathematics AP Board Solution
Exercise 12.1- 8x, 24 Find the common factors of the given terms in each.
- 3a, 21ab Find the common factors of the given terms in each.
- 7xy, 35x^2 y^3 Find the common factors of the given terms in each.…
- 4m^2 , 6m^2 , 8m^3 Find the common factors of the given terms in each.…
- 15p, 20qr, 25rp Find the common factors of the given terms in each.…
- 4x^2 , 6xy, 8y^2 x Find the common factors of the given terms in each.…
- 12x^2 y, 18xy^2 Find the common factors of the given terms in each.…
- 5x^2 - 25xy Factorise the following expressions
- 9a^2 - 6ax Factorise the following expressions
- 7p^2 + 49pq Factorise the following expressions
- 36a^2 b - 60 a^2 bc Factorise the following expressions
- 3a^2 bc + 6ab^2 c + 9abc^2 Factorise the following expressions
- 4p^2 + 5pq - 6pq^2 Factorise the following expressions
- ut + at^2 Factorise the following expressions
- 3ax - 6xy + 8by - 4ab Factorise the following:
- x^3 + 2x^2 + 5x + 10 Factorise the following:
- m^2 - mn + 4m - 4n Factorise the following:
- a^3 - a^2 b^2 - ab + b^3 Factorise the following:
- p^2 q - pr^2 - pq + r^2 Factorise the following:
Exercise 12.2- a^2 + 10a + 25 Factories the following expression-
- l^2 - 16l + 64 Factories the following expression-
- 36x^2 + 96xy + 64y^2 Factories the following expression-
- 25x^2 + 9y^2 - 30xy Factories the following expression-
- 25m^2 - 40mn + 16n^2 Factories the following expression-
- 81x^2 - 198 xy + 121y^2 Factories the following expression-
- (x + y)^2 - 4xy (Hint: first expand (x + y)^2 Factories the following…
- l^4 + 4l^2 m^2 + 4m^4 Factories the following expression-
- x^2 - 36 Factories the following
- 49x^2 - 25y^2 Factories the following
- m^2 - 121 Factories the following
- 81 - 64x^2 Factories the following
- x^2 y^2 - 64 Factories the following
- 6x^2 - 54 Factories the following
- x^2 - 81 Factories the following
- 2x - 32x^5 Factories the following
- 81x^4 - 121x^2 Factories the following
- (p^2 - 2pq + q^2) - r^2 Factories the following
- (x + y)^2 - (x - y)^2 Factories the following
- lx^2 + mx Factories the expressions-
- 7y^2 + 35z^2 Factories the expressions-
- 3x^4 + 6x^3 y + 9x^2 z Factories the expressions-
- x^2 - ax - bx + ab Factories the expressions-
- 3ax - 6ay - 8by + 4bx Factories the expressions-
- mn + m + n + 1 Factories the expressions-
- 6ab - b^2 + 12ac - 2bc Factories the expressions-
- p^2 q - pr^2 - pq + r^2 Factories the expressions-
- x (y + z) - 5 (y + z) Factories the expressions-
- x^4 - y^4 Factories the following
- a^4 - (b + c)^4 Factories the following
- l^2 - (m - n)^2 Factories the following
- 49x^2 - 16/25 Factories the following
- x^4 - 2x^2 y^2 + y^4 Factories the following
- 4 (a + b)^2 - 9 (a - b)^2 Factories the following
- a^2 + 10a + 24 Factories the following expressions
- x^2 + 9x + 18 Factories the following expressions
- p^2 - 10p + 21 Factories the following expressions
- x^2 - 4x - 32 Factories the following expressions
- The lengths of the sides of a triangle are integrals, and its area is also…
- Find the values of ‘m’ for which x^2 + 3xy + x + my -m has two linear factors…
Exercise 12.3- 48a^3 by 6a Carry out the following divisions
- 14x^3 by 42x^2 Carry out the following divisions
- 72a^3 b^4 c^5 by 8ab^2 c^3 Carry out the following divisions
- 11xy^2 z^3 by 55xyz Carry out the following divisions
- -54l^4 m^3 n^2 by 9l^2 m^2 n^2 Carry out the following divisions
- (3x^2 - 2x) ÷ x Divide the given polynomial by the given monomial…
- (5a^3 b - 7ab^3) ÷ ab Divide the given polynomial by the given monomial…
- (25x^5 - 15x^4) ÷ 5x^3 Divide the given polynomial by the given monomial…
- 4(l^5 - 6l^4 + 8l^3) ÷ 2l^2 Divide the given polynomial by the given monomial…
- 15 (a^3 b^2 c^2 - a^2 b^3 c^2 + a^2 b^2 c^3) ÷ 3abc Divide the given…
- (3p^3 - 9p^2 q - 6pq^2) ÷ (-3p) Divide the given polynomial by the given…
- (2/3 a^2 b^2 c^2 + 4/3 ab^2 c^2) ÷ 1/2 abc Divide the given polynomial by the…
- (49x - 63) ÷ 7 Workout the following divisions:
- 12x (8x - 20) ÷ 4(2x - 5) Workout the following divisions:
- 11a^3 b^3 (7c - 35) ÷ 3a^2 b^2 (c - 5) Workout the following divisions:…
- 54lmn (l + m) (m + n) (n + l) ÷ 81mn (l + m) (n + l) Workout the following…
- 36 (x + 4) (x^2 + 7x + 10) ÷ 9 (x + 4) Workout the following divisions:…
- a (a + 1) (a + 2) (a + 3) ÷ a (a + 3) Workout the following divisions:…
- (x^2 + 7x + 12) ÷ (x + 3) Factorize the expressions and divide them as…
- (x^2 - 8x + 12) ÷ (x - 6) Factorize the expressions and divide them as…
- (p^2 + 5p + 4) ÷ (p + 1) Factorize the expressions and divide them as…
- 15ab (a^2 -7a + 10) ÷ 3b (a - 2) Factorize the expressions and divide them as…
- 15lm (2p^2 -2q^2) ÷ 3l (p + q) Factorize the expressions and divide them as…
- 26z^3 (32z^2 -18) ÷ 13z^2 (4z - 3) Factorize the expressions and divide them…
Exercise 12.4- 3(x - 9) = 3x - 9 Find the errors and correct the following mathematical…
- x(3x + 2) = 3x^2 + 2 Find the errors and correct the following mathematical…
- 2x + 3x = 5x^2 Find the errors and correct the following mathematical…
- 2x + x + 3x = 5x Find the errors and correct the following mathematical…
- 4p + 3p + 2p + p - 9p = 0 Find the errors and correct the following…
- 3x + 2y = 6xy Find the errors and correct the following mathematical sentences…
- (3x)^2 + 4x + 7 = 3x^2 + 4x + 7 Find the errors and correct the following…
- (2x)^2 + 5x = 4x + 5x = 9x Find the errors and correct the following…
- (2a + 3)^2 = 2a^2 + 6a + 9 Find the errors and correct the following…
- Substitute x = - 3 in (a) x^2 + 7x + 12 = (-3)^2 + 7 (-3) + 12 = 9 + 4 + 12 =…
- Substitute x = - 3 in (b) x^2 - 5x + 6 = (-3)^2 -5 (-3) + 6 = 9 - 15 + 6 = 0…
- Substitute x = - 3 in (c) x^2 + 5x = (-3)^2 + 5 (-3) + 6 = - 9 - 15 = -24 Find…
- (x - 4)^2 = x^2 - 16 Find the errors and correct the following mathematical…
- (x + 7)^2 = x^2 + 49 Find the errors and correct the following mathematical…
- (3a + 4b) (a - b) = 3a^2 - 4a^2 Find the errors and correct the following…
- (x + 4) (x + 2) = x^2 + 8 Find the errors and correct the following…
- (x - 4) (x - 2) = x^2 - 8 Find the errors and correct the following…
- 5x^3 ÷ 5x^3 = 0 Find the errors and correct the following mathematical…
- 2x^3 + 1 ÷ 2x^3 = 1 Find the errors and correct the following mathematical…
- 3x + 2 ÷ 3x = 2/3x Find the errors and correct the following mathematical…
- 3x + 5 ÷ 3 = 5 Find the errors and correct the following mathematical…
- 4x+3/3 = x + 1 Find the errors and correct the following mathematical…
- 8x, 24 Find the common factors of the given terms in each.
- 3a, 21ab Find the common factors of the given terms in each.
- 7xy, 35x^2 y^3 Find the common factors of the given terms in each.…
- 4m^2 , 6m^2 , 8m^3 Find the common factors of the given terms in each.…
- 15p, 20qr, 25rp Find the common factors of the given terms in each.…
- 4x^2 , 6xy, 8y^2 x Find the common factors of the given terms in each.…
- 12x^2 y, 18xy^2 Find the common factors of the given terms in each.…
- 5x^2 - 25xy Factorise the following expressions
- 9a^2 - 6ax Factorise the following expressions
- 7p^2 + 49pq Factorise the following expressions
- 36a^2 b - 60 a^2 bc Factorise the following expressions
- 3a^2 bc + 6ab^2 c + 9abc^2 Factorise the following expressions
- 4p^2 + 5pq - 6pq^2 Factorise the following expressions
- ut + at^2 Factorise the following expressions
- 3ax - 6xy + 8by - 4ab Factorise the following:
- x^3 + 2x^2 + 5x + 10 Factorise the following:
- m^2 - mn + 4m - 4n Factorise the following:
- a^3 - a^2 b^2 - ab + b^3 Factorise the following:
- p^2 q - pr^2 - pq + r^2 Factorise the following:
- a^2 + 10a + 25 Factories the following expression-
- l^2 - 16l + 64 Factories the following expression-
- 36x^2 + 96xy + 64y^2 Factories the following expression-
- 25x^2 + 9y^2 - 30xy Factories the following expression-
- 25m^2 - 40mn + 16n^2 Factories the following expression-
- 81x^2 - 198 xy + 121y^2 Factories the following expression-
- (x + y)^2 - 4xy (Hint: first expand (x + y)^2 Factories the following…
- l^4 + 4l^2 m^2 + 4m^4 Factories the following expression-
- x^2 - 36 Factories the following
- 49x^2 - 25y^2 Factories the following
- m^2 - 121 Factories the following
- 81 - 64x^2 Factories the following
- x^2 y^2 - 64 Factories the following
- 6x^2 - 54 Factories the following
- x^2 - 81 Factories the following
- 2x - 32x^5 Factories the following
- 81x^4 - 121x^2 Factories the following
- (p^2 - 2pq + q^2) - r^2 Factories the following
- (x + y)^2 - (x - y)^2 Factories the following
- lx^2 + mx Factories the expressions-
- 7y^2 + 35z^2 Factories the expressions-
- 3x^4 + 6x^3 y + 9x^2 z Factories the expressions-
- x^2 - ax - bx + ab Factories the expressions-
- 3ax - 6ay - 8by + 4bx Factories the expressions-
- mn + m + n + 1 Factories the expressions-
- 6ab - b^2 + 12ac - 2bc Factories the expressions-
- p^2 q - pr^2 - pq + r^2 Factories the expressions-
- x (y + z) - 5 (y + z) Factories the expressions-
- x^4 - y^4 Factories the following
- a^4 - (b + c)^4 Factories the following
- l^2 - (m - n)^2 Factories the following
- 49x^2 - 16/25 Factories the following
- x^4 - 2x^2 y^2 + y^4 Factories the following
- 4 (a + b)^2 - 9 (a - b)^2 Factories the following
- a^2 + 10a + 24 Factories the following expressions
- x^2 + 9x + 18 Factories the following expressions
- p^2 - 10p + 21 Factories the following expressions
- x^2 - 4x - 32 Factories the following expressions
- The lengths of the sides of a triangle are integrals, and its area is also…
- Find the values of ‘m’ for which x^2 + 3xy + x + my -m has two linear factors…
- 48a^3 by 6a Carry out the following divisions
- 14x^3 by 42x^2 Carry out the following divisions
- 72a^3 b^4 c^5 by 8ab^2 c^3 Carry out the following divisions
- 11xy^2 z^3 by 55xyz Carry out the following divisions
- -54l^4 m^3 n^2 by 9l^2 m^2 n^2 Carry out the following divisions
- (3x^2 - 2x) ÷ x Divide the given polynomial by the given monomial…
- (5a^3 b - 7ab^3) ÷ ab Divide the given polynomial by the given monomial…
- (25x^5 - 15x^4) ÷ 5x^3 Divide the given polynomial by the given monomial…
- 4(l^5 - 6l^4 + 8l^3) ÷ 2l^2 Divide the given polynomial by the given monomial…
- 15 (a^3 b^2 c^2 - a^2 b^3 c^2 + a^2 b^2 c^3) ÷ 3abc Divide the given…
- (3p^3 - 9p^2 q - 6pq^2) ÷ (-3p) Divide the given polynomial by the given…
- (2/3 a^2 b^2 c^2 + 4/3 ab^2 c^2) ÷ 1/2 abc Divide the given polynomial by the…
- (49x - 63) ÷ 7 Workout the following divisions:
- 12x (8x - 20) ÷ 4(2x - 5) Workout the following divisions:
- 11a^3 b^3 (7c - 35) ÷ 3a^2 b^2 (c - 5) Workout the following divisions:…
- 54lmn (l + m) (m + n) (n + l) ÷ 81mn (l + m) (n + l) Workout the following…
- 36 (x + 4) (x^2 + 7x + 10) ÷ 9 (x + 4) Workout the following divisions:…
- a (a + 1) (a + 2) (a + 3) ÷ a (a + 3) Workout the following divisions:…
- (x^2 + 7x + 12) ÷ (x + 3) Factorize the expressions and divide them as…
- (x^2 - 8x + 12) ÷ (x - 6) Factorize the expressions and divide them as…
- (p^2 + 5p + 4) ÷ (p + 1) Factorize the expressions and divide them as…
- 15ab (a^2 -7a + 10) ÷ 3b (a - 2) Factorize the expressions and divide them as…
- 15lm (2p^2 -2q^2) ÷ 3l (p + q) Factorize the expressions and divide them as…
- 26z^3 (32z^2 -18) ÷ 13z^2 (4z - 3) Factorize the expressions and divide them…
- 3(x - 9) = 3x - 9 Find the errors and correct the following mathematical…
- x(3x + 2) = 3x^2 + 2 Find the errors and correct the following mathematical…
- 2x + 3x = 5x^2 Find the errors and correct the following mathematical…
- 2x + x + 3x = 5x Find the errors and correct the following mathematical…
- 4p + 3p + 2p + p - 9p = 0 Find the errors and correct the following…
- 3x + 2y = 6xy Find the errors and correct the following mathematical sentences…
- (3x)^2 + 4x + 7 = 3x^2 + 4x + 7 Find the errors and correct the following…
- (2x)^2 + 5x = 4x + 5x = 9x Find the errors and correct the following…
- (2a + 3)^2 = 2a^2 + 6a + 9 Find the errors and correct the following…
- Substitute x = - 3 in (a) x^2 + 7x + 12 = (-3)^2 + 7 (-3) + 12 = 9 + 4 + 12 =…
- Substitute x = - 3 in (b) x^2 - 5x + 6 = (-3)^2 -5 (-3) + 6 = 9 - 15 + 6 = 0…
- Substitute x = - 3 in (c) x^2 + 5x = (-3)^2 + 5 (-3) + 6 = - 9 - 15 = -24 Find…
- (x - 4)^2 = x^2 - 16 Find the errors and correct the following mathematical…
- (x + 7)^2 = x^2 + 49 Find the errors and correct the following mathematical…
- (3a + 4b) (a - b) = 3a^2 - 4a^2 Find the errors and correct the following…
- (x + 4) (x + 2) = x^2 + 8 Find the errors and correct the following…
- (x - 4) (x - 2) = x^2 - 8 Find the errors and correct the following…
- 5x^3 ÷ 5x^3 = 0 Find the errors and correct the following mathematical…
- 2x^3 + 1 ÷ 2x^3 = 1 Find the errors and correct the following mathematical…
- 3x + 2 ÷ 3x = 2/3x Find the errors and correct the following mathematical…
- 3x + 5 ÷ 3 = 5 Find the errors and correct the following mathematical…
- 4x+3/3 = x + 1 Find the errors and correct the following mathematical…
Exercise 12.1
Question 1.Find the common factors of the given terms in each.
8x, 24
Answer:∴ Given terms are 8x and 24
Prime factors of given terms are:-
8x = 2 × 2 × 2 × x
24 = 2 × 2 × 2 × 3
As the x is an undefined value,
Common factors will be
2 × 2 × 2 = 8
Question 2.Find the common factors of the given terms in each.
3a, 21ab
Answer:∴ Given terms are 3a and 21ab
Prime factors of given terms are:-
3a = 3 × a
21ab = a × b × 7 × 3
Common factors will be
⇒ 3 × a = 3a
Question 3.Find the common factors of the given terms in each.
7xy, 35x2y3
Answer:∴ Given terms are 7xy and 35x2y3
Prime factors of given terms are:-
7xy = 7 × x × y
35x2y3 = x × x × y × y × 7 × 5
Common factors will be
⇒ 7 × x × y = 7xy
Question 4.Find the common factors of the given terms in each.
4m2, 6m2, 8m3
Answer:∴ Given terms are 4m2,6m2 and 8m3
Prime factors of given terms are:-
4m2 = 2 × 2 × m × m
6m2 = 3 × 2 × m × m
8m3 = 2 × 2 × 2 × m × m × m
Common factors will be
⇒ 2 × m × m = 2m2
Question 5.Find the common factors of the given terms in each.
15p, 20qr, 25rp
Answer:∴ Given terms are 15p,20qr and 25rp
Prime factors of given terms are:-
15p = 3 × 5 × p
20qr = 2 × 2 × 5 × q × r
25rp = 5 × 5 × r × p
⇒ Common factors will be 5
Question 6.Find the common factors of the given terms in each.
4x2, 6xy, 8y2x
Answer:∴ Given terms are 4x2,6xy and 8y2x
Prime factors of given terms are:-
4x2 = 2 × 2 × x × x
6xy = 3 × 2 × x × y
8y2x = 2 × 2 × 2 × y × y × x
Common factors will be
⇒ 2 × x = 2x
Question 7.Find the common factors of the given terms in each.
12x2y, 18xy2
Answer:Given terms are 12x2yand 18xy2
Prime factors of given terms are:-
12yx2 = 3 × 2 × 2 × y × x × x
18xy2 = 3 × 2 × 3 × x × y × y
Common factors will be
⇒ 2 × 3 × x × y = 6xy
Question 8.Factorise the following expressions
5x2 – 25xy
Answer:In the given expression
Check the common factors for all terms;
⇒ [5 × x × x - 5 × 5 × x × y]
⇒ 5 × x[x-5 × y]
⇒ 5x[x-5y]
∴ 5x2 - 25xy = 5x[x-5y]
Question 9.Factorise the following expressions
9a2 – 6ax
Answer:In the given expression
Check the common factors for all terms;
⇒ [5 × a × a- 2 × 3 × x × a]
⇒ a[5 × a-2 × 3 × x]
⇒ a[5a-6x]
∴ 9a2 - 6ax = a[5a-6x]
Question 10.Factorise the following expressions
7p2 + 49pq
Answer:In the given expression
Check the common factors for all terms;
⇒ [7 × p × p + 7 × 7 × p × q]
⇒ 7 × p[p + 7 × q]
⇒ 7p[p + 7q]
∴ 7p2 + 49pq = 7p[p + 7q]
Question 11.Factorise the following expressions
36a2b – 60 a2bc
Answer:In the given expression
Check the common factors for all terms;
⇒ [2 × 2 × 3 × 3 × a × a × b - 2 × 2 × 3 × 5 × a × a × b × c]
⇒ 2 × 2 × 3 × a × a × b[3 × b-5 × c]
⇒ 12a2b[3b-5c]
∴ 36a2b - 60 a2bc = 12a2b[3b-5c]
Question 12.Factorise the following expressions
3a2bc + 6ab2c + 9abc2
Answer:In the given expression
Check the common factors for all terms;
⇒ [3 × a × a × b × c + 2 × 3 × a × b × b × c + 3 × 3 × a × b × c × c]
⇒ 3 × a × b × c[a + 2 × b + 3 × c]
⇒ 3abc[a + 2b + 3c]
∴ 3a2bc + 6ab2c + 9abc2 = 3abc[a + 2b + 3c]
Question 13.Factorise the following expressions
4p2 + 5pq – 6pq2
Answer:In the given expression
Check the common factors for all terms;
⇒ [2 × 2 × p × p + 5 × p × q - 2 × 3 × p × q × q]
⇒ p[2 × 2 × p + 5 × q - 2 × 3 × q × q]
⇒ p[4p + 5q-6q2]
∴ 4p2 + 5pq – 6pq2 = p[4p + 5q-6q2]
Question 14.Factorise the following expressions
ut + at2
Answer:In the given expression
Check the common factors for all terms;
⇒ [u × t + a × t × t]
⇒ t[u + a × t]
⇒ t[u + at]
∴ ut + at2 = t[u + at]
Question 15.Factorise the following:
3ax – 6xy + 8by – 4ab
Answer:In the given expression
Check weather there is any common factors for all terms;
None;
Regrouping the 1st 2 terms we have,
3ax-6xy = 3x[a-2y] -------eq 1
Regrouping the last 2 terms we have,
8by-4ab = -4b[a-2y] -------eq 2
∵ we have to make common parts in both eq 1 and 2
Combining eq 1 and 2
3ax – 6xy + 8by – 4ab = 3x[a-2y] + [-4b[a-2y] ]
= 3x[a-2y] - 4b[a-2y]
= [3x-4] [a-2y]
Hence the factors of 3ax – 6xy + 8by – 4ab are [3x-4] and [a-2y]
Question 16.Factorise the following:
x3 + 2x2 + 5x + 10
Answer:In the given expression
Check whether there is any common factors for all terms;
None;
Regrouping the 1st 2 terms we have,
x3 + 2x2 = x2[x + 2] -------eq 1
Regrouping the last 2 terms we have,
5x + 10 = 5[x + 2] -------eq 2
Combining eq 1 and 2
x3 + 2x2 + 5x + 10 = x2[x + 2] + 5[x + 2]
= [x2 + 5][x + 2]
Hence the factors of x3 + 2x2 + 5x + 10 are [x2 + 5] and [x + 2]
Question 17.Factorise the following:
m2 – mn + 4m – 4n
Answer:In the given expression
Check weather there is any common factors for all terms;
None;
Regrouping the 1st 2 terms we have,
m2 - mn= m[m - n] -------eq 1
Regrouping the last 2 terms we have,
4m – 4n = 4[m – n] -------eq 2
Combining eq 1 and 2
m2 – mn + 4m – 4n = 4[m – n] + m[m - n]
= [4 + m][m-n]
Hence the factors of m2 – mn + 4m – 4n are [m – n] and [4 + m]
Question 18.Factorise the following:
a3 – a2b2 – ab + b3
Answer:In the given expression
Check weather there is any common factors for all terms;
None;
Regrouping the 1st 2 terms we have,
a3 – a2b2 = a2[a-b2] -------eq 1
Regrouping the last 2 terms we have,
– ab + b3 = -b[a-b2] -------eq 2
∵ we have to make common parts in both eq 1 and 2
Combining eq 1 and 2
a3 – a2b2 – ab + b3 = a2[a-b2] -b[a-b2]
= [a2 – b][a – b2]
Hence the factors of a3 – a2b2 – ab + b3 are [a2 – b] and[a – b2]
Question 19.Factorise the following:
p2q – pr2 – pq + r2
Answer:In the given expression
Check weather there is any common factors for all terms;
None;
Regrouping the 1st 2 terms we have,
p2q – pr2 = p[pq-r2] -------eq 1
Regrouping the last 2 terms we have,
– pq + r2 = -1[pq-r2] -------eq 2
∵ we have to make common parts in both eq 1 and 2
Combining eq 1 and 2
p2q – pr2 – pq + r2 = p[pq-r2] -1[pq-r2]
= [p – 1][pq – r2]
Hence the factors of p2q – pr2 – pq + r2 are [p – 1] and [pq – r2]
Find the common factors of the given terms in each.
8x, 24
Answer:
∴ Given terms are 8x and 24
Prime factors of given terms are:-
8x = 2 × 2 × 2 × x
24 = 2 × 2 × 2 × 3
As the x is an undefined value,
Common factors will be
2 × 2 × 2 = 8
Question 2.
Find the common factors of the given terms in each.
3a, 21ab
Answer:
∴ Given terms are 3a and 21ab
Prime factors of given terms are:-
3a = 3 × a
21ab = a × b × 7 × 3
Common factors will be
⇒ 3 × a = 3a
Question 3.
Find the common factors of the given terms in each.
7xy, 35x2y3
Answer:
∴ Given terms are 7xy and 35x2y3
Prime factors of given terms are:-
7xy = 7 × x × y
35x2y3 = x × x × y × y × 7 × 5
Common factors will be
⇒ 7 × x × y = 7xy
Question 4.
Find the common factors of the given terms in each.
4m2, 6m2, 8m3
Answer:
∴ Given terms are 4m2,6m2 and 8m3
Prime factors of given terms are:-
4m2 = 2 × 2 × m × m
6m2 = 3 × 2 × m × m
8m3 = 2 × 2 × 2 × m × m × m
Common factors will be
⇒ 2 × m × m = 2m2
Question 5.
Find the common factors of the given terms in each.
15p, 20qr, 25rp
Answer:
∴ Given terms are 15p,20qr and 25rp
Prime factors of given terms are:-
15p = 3 × 5 × p
20qr = 2 × 2 × 5 × q × r
25rp = 5 × 5 × r × p
⇒ Common factors will be 5
Question 6.
Find the common factors of the given terms in each.
4x2, 6xy, 8y2x
Answer:
∴ Given terms are 4x2,6xy and 8y2x
Prime factors of given terms are:-
4x2 = 2 × 2 × x × x
6xy = 3 × 2 × x × y
8y2x = 2 × 2 × 2 × y × y × x
Common factors will be
⇒ 2 × x = 2x
Question 7.
Find the common factors of the given terms in each.
12x2y, 18xy2
Answer:
Given terms are 12x2yand 18xy2
Prime factors of given terms are:-
12yx2 = 3 × 2 × 2 × y × x × x
18xy2 = 3 × 2 × 3 × x × y × y
Common factors will be
⇒ 2 × 3 × x × y = 6xy
Question 8.
Factorise the following expressions
5x2 – 25xy
Answer:
In the given expression
Check the common factors for all terms;
⇒ [5 × x × x - 5 × 5 × x × y]
⇒ 5 × x[x-5 × y]
⇒ 5x[x-5y]
∴ 5x2 - 25xy = 5x[x-5y]
Question 9.
Factorise the following expressions
9a2 – 6ax
Answer:
In the given expression
Check the common factors for all terms;
⇒ [5 × a × a- 2 × 3 × x × a]
⇒ a[5 × a-2 × 3 × x]
⇒ a[5a-6x]
∴ 9a2 - 6ax = a[5a-6x]
Question 10.
Factorise the following expressions
7p2 + 49pq
Answer:
In the given expression
Check the common factors for all terms;
⇒ [7 × p × p + 7 × 7 × p × q]
⇒ 7 × p[p + 7 × q]
⇒ 7p[p + 7q]
∴ 7p2 + 49pq = 7p[p + 7q]
Question 11.
Factorise the following expressions
36a2b – 60 a2bc
Answer:
In the given expression
Check the common factors for all terms;
⇒ [2 × 2 × 3 × 3 × a × a × b - 2 × 2 × 3 × 5 × a × a × b × c]
⇒ 2 × 2 × 3 × a × a × b[3 × b-5 × c]
⇒ 12a2b[3b-5c]
∴ 36a2b - 60 a2bc = 12a2b[3b-5c]
Question 12.
Factorise the following expressions
3a2bc + 6ab2c + 9abc2
Answer:
In the given expression
Check the common factors for all terms;
⇒ [3 × a × a × b × c + 2 × 3 × a × b × b × c + 3 × 3 × a × b × c × c]
⇒ 3 × a × b × c[a + 2 × b + 3 × c]
⇒ 3abc[a + 2b + 3c]
∴ 3a2bc + 6ab2c + 9abc2 = 3abc[a + 2b + 3c]
Question 13.
Factorise the following expressions
4p2 + 5pq – 6pq2
Answer:
In the given expression
Check the common factors for all terms;
⇒ [2 × 2 × p × p + 5 × p × q - 2 × 3 × p × q × q]
⇒ p[2 × 2 × p + 5 × q - 2 × 3 × q × q]
⇒ p[4p + 5q-6q2]
∴ 4p2 + 5pq – 6pq2 = p[4p + 5q-6q2]
Question 14.
Factorise the following expressions
ut + at2
Answer:
In the given expression
Check the common factors for all terms;
⇒ [u × t + a × t × t]
⇒ t[u + a × t]
⇒ t[u + at]
∴ ut + at2 = t[u + at]
Question 15.
Factorise the following:
3ax – 6xy + 8by – 4ab
Answer:
In the given expression
Check weather there is any common factors for all terms;
None;
Regrouping the 1st 2 terms we have,
3ax-6xy = 3x[a-2y] -------eq 1
Regrouping the last 2 terms we have,
8by-4ab = -4b[a-2y] -------eq 2
∵ we have to make common parts in both eq 1 and 2
Combining eq 1 and 2
3ax – 6xy + 8by – 4ab = 3x[a-2y] + [-4b[a-2y] ]
= 3x[a-2y] - 4b[a-2y]
= [3x-4] [a-2y]
Hence the factors of 3ax – 6xy + 8by – 4ab are [3x-4] and [a-2y]
Question 16.
Factorise the following:
x3 + 2x2 + 5x + 10
Answer:
In the given expression
Check whether there is any common factors for all terms;
None;
Regrouping the 1st 2 terms we have,
x3 + 2x2 = x2[x + 2] -------eq 1
Regrouping the last 2 terms we have,
5x + 10 = 5[x + 2] -------eq 2
Combining eq 1 and 2
x3 + 2x2 + 5x + 10 = x2[x + 2] + 5[x + 2]
= [x2 + 5][x + 2]
Hence the factors of x3 + 2x2 + 5x + 10 are [x2 + 5] and [x + 2]
Question 17.
Factorise the following:
m2 – mn + 4m – 4n
Answer:
In the given expression
Check weather there is any common factors for all terms;
None;
Regrouping the 1st 2 terms we have,
m2 - mn= m[m - n] -------eq 1
Regrouping the last 2 terms we have,
4m – 4n = 4[m – n] -------eq 2
Combining eq 1 and 2
m2 – mn + 4m – 4n = 4[m – n] + m[m - n]
= [4 + m][m-n]
Hence the factors of m2 – mn + 4m – 4n are [m – n] and [4 + m]
Question 18.
Factorise the following:
a3 – a2b2 – ab + b3
Answer:
In the given expression
Check weather there is any common factors for all terms;
None;
Regrouping the 1st 2 terms we have,
a3 – a2b2 = a2[a-b2] -------eq 1
Regrouping the last 2 terms we have,
– ab + b3 = -b[a-b2] -------eq 2
∵ we have to make common parts in both eq 1 and 2
Combining eq 1 and 2
a3 – a2b2 – ab + b3 = a2[a-b2] -b[a-b2]
= [a2 – b][a – b2]
Hence the factors of a3 – a2b2 – ab + b3 are [a2 – b] and[a – b2]
Question 19.
Factorise the following:
p2q – pr2 – pq + r2
Answer:
In the given expression
Check weather there is any common factors for all terms;
None;
Regrouping the 1st 2 terms we have,
p2q – pr2 = p[pq-r2] -------eq 1
Regrouping the last 2 terms we have,
– pq + r2 = -1[pq-r2] -------eq 2
∵ we have to make common parts in both eq 1 and 2
Combining eq 1 and 2
p2q – pr2 – pq + r2 = p[pq-r2] -1[pq-r2]
= [p – 1][pq – r2]
Hence the factors of p2q – pr2 – pq + r2 are [p – 1] and [pq – r2]
Exercise 12.2
Question 1.Factories the following expression-
a2 + 10a + 25
Answer:In the given expression
1st and last terms are perfect square
⇒ a2 = a × a
⇒ 25 = 5 × 5
And the middle expression is in form of 2ab
10a = 2 × 5 × a
∴ a × a + 2 × 5 × a + 5 × 5
Gives (a + b)2 = a2 + 2ab + b2
⇒ In a2 + 10a + 25
a = a and b = 5;
∴ a2 + 10a + 25 = (a + 5)2
Hence the factors of a2 + 10a + 25 are (a + 5) and (a + 5)
Question 2.Factories the following expression-
l2 – 16l + 64
Answer:In the given expression
1st and last terms are perfect square
⇒ l2 = l × l
⇒ 64 = 8 × 8
And the middle expression is in form of 2ab
16l = 2 × 8 × l
∴ l × l + 2 × 8 × l + 8 × 8
Gives (a-b)2 = a2-2ab + b2
⇒ In l2 + 16l + 64
a = l and b = 8;
∴ l2 + 16l + 64 = (l + 8)2
Hence the factors of l2 + 16l + 64 are (l + 8) and (l + 8)
Question 3.Factories the following expression-
36x2 + 96xy + 64y2
Answer:In the given expression
1st and last terms are perfect square
⇒ 36x2 = 6x × 6x
⇒ 64y2 = 8y × 8y
And the middle expression is in form of 2ab
96xy = 2 × 6x × 8y
∴ 6x × 6x + 2 × 8y × 6x + 8y × 8y
Gives (a + b)2 = a2 + 2ab + b2
⇒ In 36x2 + 96xy + 64y2
a = 6x and b = 8y;
∴ 36x2 + 96xy + 64y2 = (6x + 8y)2
Hence the factors of 36x2 + 96xy + 64y2 are (6x + 8y) and (6x + 8y)
Question 4.Factories the following expression-
25x2 + 9y2 – 30xy
Answer:In the given expression
1st and last terms are perfect square
⇒ 25x2 = 5x × 5x
⇒ 9y2 = 3y × 3y
And the middle expression is in form of 2ab
30xy = 2 × 5x × 3y
∴ 5x × 5x + 2 × 3y × 5x + 3y × 3y
Gives (a-b)2 = a2-2ab + b2
⇒ In 25x2 – 30xy + 9y2
a = 5x and b = 3y;
∴ 25x2 - 30xy + 9y2 = (5x-3y)2
Hence the factors of 25x2 - 30xy + 9y2 are (5x-3y) and (5x-3y)
Question 5.Factories the following expression-
25m2 – 40mn + 16n2
Answer:In the given expression
1st and last terms are perfect square
⇒ 25m2 = 5m × 5m
⇒ 16n2 = 4n × 4n
And the middle expression is in form of 2ab
40mn = 2 × 5m × 4n
∴ 5m × 5m - 2 × 4n × 5m + 4n × 4n
Gives (a-b)2 = a2-2ab + b2
⇒ In 25m2 – 40mn + 16n2
a = 5m and b = 4n;
∴ 25m2 – 40mn + 16n2 = (5m-4n)2
Hence the factors of 25m2 – 40mn + 16n2 are (5m-4n)and (5m-4n)
Question 6.Factories the following expression-
81x2– 198 xy + 121y2
Answer:In the given expression
1st and last terms are perfect square
⇒ 81x2 = 9x × 9x
⇒ 121y2 = 11y × 11y
And the middle expression is in form of 2ab
198xy = 2 × 9x × 11y
∴ 9x × 9x - 2 × 11y × 9x + 11y × 11y
Gives (a-b)2 = a2-2ab + b2
⇒ In 81x2 – 198xy + 121y2
a = 9x and b = 11y;
∴ 81x2 – 198xy + 121y2 = (9x-11y)2
Hence the factors of 81x2 – 198xy + 121y2 are (9x-11y)and (9x-11y)
Question 7.Factories the following expression-
(x + y)2 – 4xy
(Hint: first expand (x + y)2
Answer:If (a + b)2 = a2 + 2ab + b2
Then (x + y)2 – 4xy
= x2 + 2xy + y2-4xy
= x2 + y2-2xy
In given expression
1st and last terms are perfect square
⇒ x2 = x × x
⇒ y2 = y × y
And the middle expression is in form of 2ab
2xy = 2 × x × y
∴ x × x - 2 × y × x + y × y
Gives (a-b)2 = a2-2ab + b2
⇒ In x2 – 2xy + y2
a = x and b = y;
∴ x2 – 2xy + y2 = (x-y)2
Hence the factors of (x + y)2 – 4xy are (x-y)and (x-y)
Question 8.Factories the following expression-
l4 + 4l2m2 + 4m4
Answer:In given expression
1st and last terms are perfect square
⇒ l4 = l2 × l2
⇒ m4 = m2 × m2
And the middle expression is in form of 2ab
4l2m2 = 2 × l2 × m2
∴ l2 × l2 + 2 × m2 × l2 + m2 × m2
Gives (a + b)2 = a2 + 2ab + b2
⇒ In l4 + 4l2m2 + m4
a = l2 and b = m2;
∴ l4 – 4l2m2 + m4 = (l2-m2)2
Hence the factors of l4 + 4l2m2 + 4m4 are (l2-m2)and (l2-m2)
Question 9.Factories the following
x2 – 36
Answer:In given expression
Both terms are perfect square
⇒ x2 = x × x
⇒ 36 = 6 × 6
∴ x2-62
Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = x and b = 6;
x2 – 36 = (x + 6)(x-6)
Hence the factors of x2 – 36 are (x + 6) and (x-6)
Question 10.Factories the following
49x2 – 25y2
Answer:In given expression
Both terms are perfect square
⇒ 49x2 = 7x × 7x
⇒ 25y2 = 5y × 5y
∴ 49x2-25y2 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = 7x and b = 5y;
49x2 – 25y2 = (7x + 5y)(7x-5y)
Hence the factors of 49x2 – 25y2 are (7x + 5y) and (7x-5y)
Question 11.Factories the following
m2 – 121
Answer:In given expression
Both terms are perfect square
⇒ m2 = m × m
⇒ 121 = 11 × 11
∴ m2-121 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = m and b = 11;
m2 – 121 = (m + 11)(m-11)
Hence the factors of m2 – 121 are (m + 11) and (m-11)
Question 12.Factories the following
81 – 64x2
Answer:In given expression
Both terms are perfect square
⇒ 64x2 = 8x × 8x
⇒ 81 = 9 × 9
∴ 81-64x2 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = 9 and b = 8x;
81-64x2 = (9-8x)(9 + 8x)
Hence the factors of 81-64x2are(9-8x) and (9 + 8x)
Question 13.Factories the following
x2y2 – 64
Answer:In given expression
Both terms are perfect square
⇒ y2x2 = xy × xy
⇒ 64 = 8 × 8
∴ x2y2 – 64 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = xy and b = 8;
x2y2 – 64 = (xy-8)(xy + 8)
Hence the factors of x2y2 – 64 are (xy-8) and (xy + 8)
Question 14.Factories the following
6x2 – 54
Answer:In given expression
Take out the common factor,
[2 × 3 × x × x-2 × 3 × 3 × 3]
⇒ 2 × 3[x × x-3 × 3]
⇒ 6[x2-9]
Both terms are perfect square
⇒ x2 = x × x
⇒ 9 = 3 × 3
∴ x2– 9 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = x and b = 3;
x2– 9 = (x-3)(x + 3)
Hence the factors of 6x2 – 54 are 6,(x-3) and (x + 3)
Question 15.Factories the following
x2– 81
Answer:In given expression
Both terms are perfect square
⇒ x2 = x × x
⇒ 81 = 9 × 9
∴ x2 – 81 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = x and b = 9;
x2 – 81 = (x-9)(x + 9)
Hence the factors of x2 – 81 are (x-9) and (x-9)
Question 16.Factories the following
2x – 32x5
Answer:In given expression
Take out the common factor,
[2 × x - 2 × 2 × 2 × 2 × 2 × x × x × x × x × x]
⇒ 2 × x[1 - 2 × 2 × 2 × 2 × x × x × x × x]
⇒ 2x [1-16x4] = 2x [1-(2x)4]
⇒ In the term 1-(2x)4
= 1-(4x2)2
Both terms are perfect square
⇒ (4x2 )2 = 4x2 × 4x2
⇒ 1 = 1 × 1
∴ 1-(4x2)2 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = 1 and b = 4x2;
1-16x4 = (1-4x2)(1 + 4x2)
→ 1-4x2 = 1-(2x)2
∴ 1-4x2 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = 1 and b = 2x;
1-4x2 = (1-2x)(1 + 2x)
∴ 1-16x2 = (1-2x)(1 + 2x) (1 + 4x2)
Hence the factors of 2x – 32x5 are 2x,(1-2x),(1 + 2x) and (1 + 4x2)
Question 17.Factories the following
81x4 – 121x2
Answer:In given expression
Take out the common factor,
[3 × 3 × 3 × 3 × x × x × x × x - 11 × 11 × x × x]
⇒ x × x[3 × 3 × 3 × 3 × x × x - 11 × 11]
⇒ x2[81x2 – 121]
In expression 81x2 - 121
Both terms are perfect square
⇒ 81x2 = 9x × 9x
⇒ 121 = 11 × 11
∴ 81x2 – 121 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = 9x and b = 11;
81x2 – 121 = (9x-11)(9x + 11)
Hence the factors of 81x4 – 121x2 are x2,(9x-11) and (9x + 11)
Question 18.Factories the following
(p2 – 2pq + q2) – r2
Answer:In the given expression p2 – 2pq + q2
1st and last terms are perfect square
⇒ p2 = p × p
⇒ q2 = q × q
And the middle expression is in form of 2ab
2pq = 2 × p × q
∴ p × p - 2 × p × q + q × q
Gives (a-b)2 = a2-2ab + b2
⇒ In p2 – 2pq + q2
a = p and b = q;
∴ p2 – 2pq + q2 = (p-q)2
Now the given expression is (p-q)2– r2
Both terms are perfect square
⇒ (p-q)2 = (p-q) × (p-q)
⇒ r2 = r × r
∴ (p-q)2– r2 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = (p-q) and b = r;
(p-q)2– r2 = (p-q-r) (p-q + r)
Hence the factors of (p2 – 2pq + q2) – r2 are (p-q-r) and (p-q + r)
Question 19.Factories the following
(x + y)2 – (x – y)2
Answer:In the given expression
We know that
(a + b)2 = a2 + 2ab + b2
(a-b)2 = a2-2ab + b2
Hence
If a = x and b = y
(x + y)2 – (x – y)2 = x2 + y2 + 2xy – [x2 + y2-2xy]
= x2 + y2 + 2xy -x2-y2 + 2xy
= 4xy
Question 20.Factories the expressions-
lx2 + mx
Answer:In the given expression
Take out the common in all the terms,
⇒ lx2 + mx
⇒ x[lx + m]
Question 21.Factories the expressions-
7y2 + 35z2
Answer:In the given expression
Take out the common in all the terms,
⇒ 7y2 + 35z2
⇒ 7[y2 + 5z2]
Question 22.Factories the expressions-
3x4 + 6x3y + 9x2z
Answer:In the given expression
Take out the common in all the terms,
⇒ 3x4 + 6x3y + 9x2z
⇒ 3x2[x2 + 2xy + 3z]
Question 23.Factories the expressions-
x2 – ax – bx + ab
Answer:In the given expression
Check weather there is any common factors for all terms;
None;
Regrouping the 1st 2 terms we have,
x2 - ax= x[x - a] -------eq 1
Regrouping the last 2 terms we have,
-bx + ab = -b[x – a] -------eq 2
Combining eq 1 and 2
x2 – ax – bx + ab = x[x - a] - b[x – a]
= [x - b][x - a]
Hence the factors of[ x2 – ax – bx + ab] are [x - b]and [x - a]
Question 24.Factories the expressions-
3ax – 6ay – 8by + 4bx
Answer:In the given expression
Check weather there is any common factors for all terms;
None;
Regrouping the 1st 2 terms we have,
3ax – 6ay= 3a[x - 2y] -------eq 1
Regrouping the last 2 terms we have,
-8by + 4bx = 4b[x – 2y] -------eq 2
Combining eq 1 and 2
3ax – 6ay – 8by + 4bx = 3a[x - 2y] + 4b[x – 2y]
= [x – 2y][3a + 4b]
Hence the factors of[3ax – 6ay – 8by + 4bx] are [x – 2y] and [3a + 4b]
Question 25.Factories the expressions-
mn + m + n + 1
Answer:In the given expression
Check weather there is any common factors for all terms;
None;
Regrouping the 1st 2 terms we have,
mn + m = m[n + 1] -------eq 1
Regrouping the last 2 terms we have,
n + 1 = 1[n + 1] -------eq 2
Combining eq 1 and 2
mn + m + n + 1 = m[n + 1] + 1[n + 1]
= [m + 1][n + 1]
Hence the factors of[mn + m + n + 1] are [m + 1] and [n + 1]
Question 26.Factories the expressions-
6ab – b2 + 12ac – 2bc
Answer:In the given expression
Check weather there is any common factors for all terms;
None;
Regrouping the 1st 2 terms we have,
6ab – b2 = b[6a - b] -------eq 1
Regrouping the last 2 terms we have,
12ac – 2bc = 2c[6a - b] -------eq 2
Combining eq 1 and 2
6ab – b2 + 12ac – 2bc = b[6a - b] + 2c[6a - b]
= [6a - b][b + 2c]
Hence the factors of[6ab – b2 + 12ac – 2bc] are [6a - b] and[b + 2c]
Question 27.Factories the expressions-
p2q – pr2 – pq + r2
Answer:In the given expression
Check weather there is any common factors for all terms;
None;
Regrouping the 1st 2 terms we have,
p2q – pr2 = p[pq – r2] -------eq 1
Regrouping the last 2 terms we have,
– pq + r2 = -1[pq – r2] -------eq 2
∵ we have to make common parts in both eq 1 and 2
Combining eq 1 and 2
p2q – pr2 – pq + r2 = p[pq – r2] -1[pq – r2]
= [pq – r2][p - 1]
Hence the factors of[p2q – pr2 – pq + r2] are [pq – r2] and [p - 1]
Question 28.Factories the expressions-
x (y + z) – 5 (y + z)
Answer:In the given expression
Take out the common in all the terms,
⇒ x (y + z) – 5 (y + z)
⇒ (y + z)(x - 5)
Hence the factors of x (y + z) – 5 (y + z) are (y + z) and (x - 5)
Question 29.Factories the following
x4 – y4
Answer:In expression x4 – y4
Both terms are perfect square
⇒ x4 = x2 × x2
⇒ y4 = y2 × y2
∴ x4 – y4 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = x2 and b = y2;
x4 – y4 = (x2 – y2)( x2 + y2),
∴ x2 – y2 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = x and b = y;
x2 – y2 = (x– y)( x+ y),
⇒ x4 – y4 = (x– y)( x+ y), ( x2 + y2)
Hence the factors of x4 – y4 are (x– y),( x+ y) and ( x2 + y2)
Question 30.Factories the following
a4 – (b + c)4
Answer:In expression a4 – (b + c)4
Both terms are perfect square
⇒ a4 = a2 × a2
⇒ (b + c)4 = (b + c)2 × (b + c)2
∴ a4 – (b + c)4 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = a2 and b = (b + c)2;
a4 – (b + c)4 = (a2 – (b + c)2)( a2 + (b + c)2),
∴ a2 – (b + c)2 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = a and b = (b + c);
a2 – (b + c)2 = (a– (b + c))( a+ (b + c)),
⇒ a4 – (b + c)4 = (a– (b + c))( a+ (b + c)), ( a2 + (b + c)2)
⇒ a4 – (b + c)4 = (a–b–c)(a + b + c), (a2 + b2 + c2 + 2bc)
Hence the factors of a4 – (b + c)4 are (a–b–c),(a + b + c),( a2 + b2 + c2 + 2bc)
Question 31.Factories the following
l2 – (m – n)2
Answer:In the given expression l2 – (m – n)2
Both terms are perfect square
⇒ l2 = l × l
⇒ (m – n)2 = (m – n) × (m – n)
∴ l2 – (m - n)2 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = a and b = (m - n);
∴ l2 – (m – n)2 = (l + m-n)(l-m + n)
Hence the factors of l2–(m–n)2are (l + m-n)(l-m + n)
Question 32.
Answer:In the given expression 49x2 –![](data:image/png;base64,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)
Both terms are perfect square
⇒ 49x2 = 7x × 7x
⇒ (
)2 = ![](data:image/png;base64,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)
49x2 –
Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = 7x and b =
;
∴ (7x)2 –(
)2 = ( 7x –
) (7x +
)
Hence the factors of 49x2 –
are ( 7x –
) and (7x +
)
Question 33.Factories the following
x4 – 2x2y2 + y4
Answer:In the given expression
1st and last terms are perfect square
⇒ x4 = x2 × x2
⇒ y4 = y2 × y2
And the middle expression is in form of 2ab
2x2y2 = 2 × x2 × y2
∴ x2 × x2 - 2 × x2 × y2 + y2 × y2
Gives (a-b)2 = a2-2ab + b2
⇒ x4 – 2x2y2 + y4
a = x2 and b = y2;
∴ x4 – 2x2y2 + y4 = (x2 – y2) (x2 + y2)
Hence the factors of x4 – 2x2y2 + y4 are (x2 – y2) and (x2 + y2)
Question 34.Factories the following
4 (a + b)2 – 9 (a – b)2
Answer:In the given expression
We know that
(a + b)2 = a2 + 2ab + b2
(a-b)2 = a2-2ab + b2
Hence
4[a2 + 2ab + b2] – 9[a2-2ab + b2]
4a2 + 8ab + 4b2 - 9a2 + 18ab - 9b2
26ab – 5a2 - 5b2
25ab + ab – 5a2 – 5b2
[25ab – 5a2] + [ab – 5b2]
5a[5b – a] – b[5b – a]
[5a – b][5b – a]
Hence the factors 4 (a + b)2 – 9 (a – b)2 are [5a – b] and [5b – a]
Question 35.Factories the following expressions
a2 + 10a + 24
Answer:The given expression looks as
x2 + (a + b)x + ab
where a + b = 10; and ab = 24;
factors of 24 their sum
1 × 24 1 + 24 = 25
12 × 2 2 + 12 = 14
6 × 4 6 + 4 = 10
∴ the factors having sum 10 are 6 and 4
a2 + 10a + 24 = a2 + (6 + 4)a + 24
= a2 + 6a + 4a + 24
= a(a + 6) + 4(a + 6)
= (a + 6)(a + 4)
Hence the factors of a2 + 10a + 24 are (a + 6) and (a + 4)
Question 36.Factories the following expressions
x2 + 9x + 18
Answer:The given expression looks as
x2 + (a + b)x + ab
where a + b = 9; and ab = 18;
factors of 18 their sum
1 × 18 1 + 18 = 19
9 × 2 2 + 9 = 11
6 × 3 6 + 3 = 9
∴ the factors having sum 9 are 6 and 3
x2 + 9x + 18 = x2 + (6 + 3)x + 18
= x2 + 6x + 3x + 18
= x(x + 6) + 3(x + 6)
= (x + 6)(x + 3)
Hence the factors of x2 + 9x + 18 are (x + 6) and (x + 3)
Question 37.Factories the following expressions
p2 – 10p + 21
Answer:The given expression looks as
x2 + (a + b)x + ab
where a + b = -10; and ab = 21;
factors of 21 their sum
-1 × -21 -1-18 = -19
-7 × -3 -7-3 = -10
∴ the factors having sum -10 are -7 and -3
p2 + 9p + 18 = p2 + (-7-3)p + 21
= p2 -7p-3p + 21
= p(p-7) -3(p-7)
= (p-7)(p-3)
Hence the factors of p2 + 9p + 18 are (p-7) and (p-3)
Question 38.Factories the following expressions
x2 – 4x – 32
Answer:The given expression looks as
x2 + (a + b)x + ab
where a + b = -4; and ab = -32;
factors of -32 their sum
1 × -32 1-32 = -31
-16 × 2 2 -16 = - 14
-8 × 4 4 -8 = -4
∴ the factors having sum -4 are -8 and 4
x2 – 4x – 32 = x2 + (4 -8)x - 32
= x2 + 4x - 8x - 32
= x(x + 4) -8(x + 4)
= (x + 4)(x-8)
Hence the factors of x2 – 4x – 32 are (x + 4) and (x-8)
Question 39.The lengths of the sides of a triangle are integrals, and its area is also integer. One side is 21 and the perimeter is 48. Find the shortest side.
Answer:A = √ s(s-a)(s-b)(s-c)
If the area is an integer
Then [s(s-a)(s-b)(s-c)] should be proper square
If s =
Then s =
= 24
Hence ;
A = √ 24(24-a)(24-b)(24-c)
If side of triangle are
a = 21 and b + c = 27
let c be smallest side
then b = 27-c
∴ √ 24(24-21)(24-27 + c)(24-c)
⇒ √ 24 × 3 × (c-3)(24-c)
⇒ √ 2 × 2 × 2 × 3 × 3 × (c-3)(24-c)
⇒ 2 × 3√2(c-3)(24-c)
⇒ 6√2(c-3)(24-c)
∴ the value of [2(c-3)(24-c)] must be a perfect square for area to be a integer
For getting square 2(c-3) should be equal to (24-c)
2(c-3) = (24-c)
2c-6 = 24-c
2c + c = 24 + 6
3c = 10
c = 10; b = 27-c = 27-10 = 17
Hence the size of smallest size is 10.
Question 40.Find the values of ‘m’ for which x2 + 3xy + x + my –m has two linear factors in x and y, with integer coefficients.
Answer:For the given 2 degree equation
That must be equal to(ax + by + c)(dx + e)
= ad.x2 + bd.xy + cd.x + ea.x + be.y + ec
= ad.x2 + bd.xy + (cd + ea).x + be.y + ec
x2 + 3xy + x + my–m = ad.x2 + bd.xy + (cd + ea).x + be.y + ec
compare the equation
and take out the coefficient of every term
a.d = 1 ----------1
b.d = 3 ----------2
c.d + e.a = 1 ----------3
b.e = m ----------4
e.c = -m ----------5
⇒ from eq 1; a = d = 1 ∵ all coefficient are integers
After putting result in eq 3; c + e = 1 -------6
After putting result in eq 2; b = 3 --------7
⇒ divide eq 4 and 5
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![](data:image/png;base64,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)
∴ that implies b = -c = -3 ∵ eq 7
Put value of c in eq 6
-3 + e = 1
e = 1 + 3 = 4
Putting value of b and e in eq 4
m = b × e
m = 3 × 4 = 12
Factories the following expression-
a2 + 10a + 25
Answer:
In the given expression
1st and last terms are perfect square
⇒ a2 = a × a
⇒ 25 = 5 × 5
And the middle expression is in form of 2ab
10a = 2 × 5 × a
∴ a × a + 2 × 5 × a + 5 × 5
Gives (a + b)2 = a2 + 2ab + b2
⇒ In a2 + 10a + 25
a = a and b = 5;
∴ a2 + 10a + 25 = (a + 5)2
Hence the factors of a2 + 10a + 25 are (a + 5) and (a + 5)
Question 2.
Factories the following expression-
l2 – 16l + 64
Answer:
In the given expression
1st and last terms are perfect square
⇒ l2 = l × l
⇒ 64 = 8 × 8
And the middle expression is in form of 2ab
16l = 2 × 8 × l
∴ l × l + 2 × 8 × l + 8 × 8
Gives (a-b)2 = a2-2ab + b2
⇒ In l2 + 16l + 64
a = l and b = 8;
∴ l2 + 16l + 64 = (l + 8)2
Hence the factors of l2 + 16l + 64 are (l + 8) and (l + 8)
Question 3.
Factories the following expression-
36x2 + 96xy + 64y2
Answer:
In the given expression
1st and last terms are perfect square
⇒ 36x2 = 6x × 6x
⇒ 64y2 = 8y × 8y
And the middle expression is in form of 2ab
96xy = 2 × 6x × 8y
∴ 6x × 6x + 2 × 8y × 6x + 8y × 8y
Gives (a + b)2 = a2 + 2ab + b2
⇒ In 36x2 + 96xy + 64y2
a = 6x and b = 8y;
∴ 36x2 + 96xy + 64y2 = (6x + 8y)2
Hence the factors of 36x2 + 96xy + 64y2 are (6x + 8y) and (6x + 8y)
Question 4.
Factories the following expression-
25x2 + 9y2 – 30xy
Answer:
In the given expression
1st and last terms are perfect square
⇒ 25x2 = 5x × 5x
⇒ 9y2 = 3y × 3y
And the middle expression is in form of 2ab
30xy = 2 × 5x × 3y
∴ 5x × 5x + 2 × 3y × 5x + 3y × 3y
Gives (a-b)2 = a2-2ab + b2
⇒ In 25x2 – 30xy + 9y2
a = 5x and b = 3y;
∴ 25x2 - 30xy + 9y2 = (5x-3y)2
Hence the factors of 25x2 - 30xy + 9y2 are (5x-3y) and (5x-3y)
Question 5.
Factories the following expression-
25m2 – 40mn + 16n2
Answer:
In the given expression
1st and last terms are perfect square
⇒ 25m2 = 5m × 5m
⇒ 16n2 = 4n × 4n
And the middle expression is in form of 2ab
40mn = 2 × 5m × 4n
∴ 5m × 5m - 2 × 4n × 5m + 4n × 4n
Gives (a-b)2 = a2-2ab + b2
⇒ In 25m2 – 40mn + 16n2
a = 5m and b = 4n;
∴ 25m2 – 40mn + 16n2 = (5m-4n)2
Hence the factors of 25m2 – 40mn + 16n2 are (5m-4n)and (5m-4n)
Question 6.
Factories the following expression-
81x2– 198 xy + 121y2
Answer:
In the given expression
1st and last terms are perfect square
⇒ 81x2 = 9x × 9x
⇒ 121y2 = 11y × 11y
And the middle expression is in form of 2ab
198xy = 2 × 9x × 11y
∴ 9x × 9x - 2 × 11y × 9x + 11y × 11y
Gives (a-b)2 = a2-2ab + b2
⇒ In 81x2 – 198xy + 121y2
a = 9x and b = 11y;
∴ 81x2 – 198xy + 121y2 = (9x-11y)2
Hence the factors of 81x2 – 198xy + 121y2 are (9x-11y)and (9x-11y)
Question 7.
Factories the following expression-
(x + y)2 – 4xy
(Hint: first expand (x + y)2
Answer:
If (a + b)2 = a2 + 2ab + b2
Then (x + y)2 – 4xy
= x2 + 2xy + y2-4xy
= x2 + y2-2xy
In given expression
1st and last terms are perfect square
⇒ x2 = x × x
⇒ y2 = y × y
And the middle expression is in form of 2ab
2xy = 2 × x × y
∴ x × x - 2 × y × x + y × y
Gives (a-b)2 = a2-2ab + b2
⇒ In x2 – 2xy + y2
a = x and b = y;
∴ x2 – 2xy + y2 = (x-y)2
Hence the factors of (x + y)2 – 4xy are (x-y)and (x-y)
Question 8.
Factories the following expression-
l4 + 4l2m2 + 4m4
Answer:
In given expression
1st and last terms are perfect square
⇒ l4 = l2 × l2
⇒ m4 = m2 × m2
And the middle expression is in form of 2ab
4l2m2 = 2 × l2 × m2
∴ l2 × l2 + 2 × m2 × l2 + m2 × m2
Gives (a + b)2 = a2 + 2ab + b2
⇒ In l4 + 4l2m2 + m4
a = l2 and b = m2;
∴ l4 – 4l2m2 + m4 = (l2-m2)2
Hence the factors of l4 + 4l2m2 + 4m4 are (l2-m2)and (l2-m2)
Question 9.
Factories the following
x2 – 36
Answer:
In given expression
Both terms are perfect square
⇒ x2 = x × x
⇒ 36 = 6 × 6
∴ x2-62
Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = x and b = 6;
x2 – 36 = (x + 6)(x-6)
Hence the factors of x2 – 36 are (x + 6) and (x-6)
Question 10.
Factories the following
49x2 – 25y2
Answer:
In given expression
Both terms are perfect square
⇒ 49x2 = 7x × 7x
⇒ 25y2 = 5y × 5y
∴ 49x2-25y2 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = 7x and b = 5y;
49x2 – 25y2 = (7x + 5y)(7x-5y)
Hence the factors of 49x2 – 25y2 are (7x + 5y) and (7x-5y)
Question 11.
Factories the following
m2 – 121
Answer:
In given expression
Both terms are perfect square
⇒ m2 = m × m
⇒ 121 = 11 × 11
∴ m2-121 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = m and b = 11;
m2 – 121 = (m + 11)(m-11)
Hence the factors of m2 – 121 are (m + 11) and (m-11)
Question 12.
Factories the following
81 – 64x2
Answer:
In given expression
Both terms are perfect square
⇒ 64x2 = 8x × 8x
⇒ 81 = 9 × 9
∴ 81-64x2 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = 9 and b = 8x;
81-64x2 = (9-8x)(9 + 8x)
Hence the factors of 81-64x2are(9-8x) and (9 + 8x)
Question 13.
Factories the following
x2y2 – 64
Answer:
In given expression
Both terms are perfect square
⇒ y2x2 = xy × xy
⇒ 64 = 8 × 8
∴ x2y2 – 64 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = xy and b = 8;
x2y2 – 64 = (xy-8)(xy + 8)
Hence the factors of x2y2 – 64 are (xy-8) and (xy + 8)
Question 14.
Factories the following
6x2 – 54
Answer:
In given expression
Take out the common factor,
[2 × 3 × x × x-2 × 3 × 3 × 3]
⇒ 2 × 3[x × x-3 × 3]
⇒ 6[x2-9]
Both terms are perfect square
⇒ x2 = x × x
⇒ 9 = 3 × 3
∴ x2– 9 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = x and b = 3;
x2– 9 = (x-3)(x + 3)
Hence the factors of 6x2 – 54 are 6,(x-3) and (x + 3)
Question 15.
Factories the following
x2– 81
Answer:
In given expression
Both terms are perfect square
⇒ x2 = x × x
⇒ 81 = 9 × 9
∴ x2 – 81 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = x and b = 9;
x2 – 81 = (x-9)(x + 9)
Hence the factors of x2 – 81 are (x-9) and (x-9)
Question 16.
Factories the following
2x – 32x5
Answer:
In given expression
Take out the common factor,
[2 × x - 2 × 2 × 2 × 2 × 2 × x × x × x × x × x]
⇒ 2 × x[1 - 2 × 2 × 2 × 2 × x × x × x × x]
⇒ 2x [1-16x4] = 2x [1-(2x)4]
⇒ In the term 1-(2x)4
= 1-(4x2)2
Both terms are perfect square
⇒ (4x2 )2 = 4x2 × 4x2
⇒ 1 = 1 × 1
∴ 1-(4x2)2 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = 1 and b = 4x2;
1-16x4 = (1-4x2)(1 + 4x2)
→ 1-4x2 = 1-(2x)2
∴ 1-4x2 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = 1 and b = 2x;
1-4x2 = (1-2x)(1 + 2x)
∴ 1-16x2 = (1-2x)(1 + 2x) (1 + 4x2)
Hence the factors of 2x – 32x5 are 2x,(1-2x),(1 + 2x) and (1 + 4x2)
Question 17.
Factories the following
81x4 – 121x2
Answer:
In given expression
Take out the common factor,
[3 × 3 × 3 × 3 × x × x × x × x - 11 × 11 × x × x]
⇒ x × x[3 × 3 × 3 × 3 × x × x - 11 × 11]
⇒ x2[81x2 – 121]
In expression 81x2 - 121
Both terms are perfect square
⇒ 81x2 = 9x × 9x
⇒ 121 = 11 × 11
∴ 81x2 – 121 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = 9x and b = 11;
81x2 – 121 = (9x-11)(9x + 11)
Hence the factors of 81x4 – 121x2 are x2,(9x-11) and (9x + 11)
Question 18.
Factories the following
(p2 – 2pq + q2) – r2
Answer:
In the given expression p2 – 2pq + q2
1st and last terms are perfect square
⇒ p2 = p × p
⇒ q2 = q × q
And the middle expression is in form of 2ab
2pq = 2 × p × q
∴ p × p - 2 × p × q + q × q
Gives (a-b)2 = a2-2ab + b2
⇒ In p2 – 2pq + q2
a = p and b = q;
∴ p2 – 2pq + q2 = (p-q)2
Now the given expression is (p-q)2– r2
Both terms are perfect square
⇒ (p-q)2 = (p-q) × (p-q)
⇒ r2 = r × r
∴ (p-q)2– r2 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = (p-q) and b = r;
(p-q)2– r2 = (p-q-r) (p-q + r)
Hence the factors of (p2 – 2pq + q2) – r2 are (p-q-r) and (p-q + r)
Question 19.
Factories the following
(x + y)2 – (x – y)2
Answer:
In the given expression
We know that
(a + b)2 = a2 + 2ab + b2
(a-b)2 = a2-2ab + b2
Hence
If a = x and b = y
(x + y)2 – (x – y)2 = x2 + y2 + 2xy – [x2 + y2-2xy]
= x2 + y2 + 2xy -x2-y2 + 2xy
= 4xy
Question 20.
Factories the expressions-
lx2 + mx
Answer:
In the given expression
Take out the common in all the terms,
⇒ lx2 + mx
⇒ x[lx + m]
Question 21.
Factories the expressions-
7y2 + 35z2
Answer:
In the given expression
Take out the common in all the terms,
⇒ 7y2 + 35z2
⇒ 7[y2 + 5z2]
Question 22.
Factories the expressions-
3x4 + 6x3y + 9x2z
Answer:
In the given expression
Take out the common in all the terms,
⇒ 3x4 + 6x3y + 9x2z
⇒ 3x2[x2 + 2xy + 3z]
Question 23.
Factories the expressions-
x2 – ax – bx + ab
Answer:
In the given expression
Check weather there is any common factors for all terms;
None;
Regrouping the 1st 2 terms we have,
x2 - ax= x[x - a] -------eq 1
Regrouping the last 2 terms we have,
-bx + ab = -b[x – a] -------eq 2
Combining eq 1 and 2
x2 – ax – bx + ab = x[x - a] - b[x – a]
= [x - b][x - a]
Hence the factors of[ x2 – ax – bx + ab] are [x - b]and [x - a]
Question 24.
Factories the expressions-
3ax – 6ay – 8by + 4bx
Answer:
In the given expression
Check weather there is any common factors for all terms;
None;
Regrouping the 1st 2 terms we have,
3ax – 6ay= 3a[x - 2y] -------eq 1
Regrouping the last 2 terms we have,
-8by + 4bx = 4b[x – 2y] -------eq 2
Combining eq 1 and 2
3ax – 6ay – 8by + 4bx = 3a[x - 2y] + 4b[x – 2y]
= [x – 2y][3a + 4b]
Hence the factors of[3ax – 6ay – 8by + 4bx] are [x – 2y] and [3a + 4b]
Question 25.
Factories the expressions-
mn + m + n + 1
Answer:
In the given expression
Check weather there is any common factors for all terms;
None;
Regrouping the 1st 2 terms we have,
mn + m = m[n + 1] -------eq 1
Regrouping the last 2 terms we have,
n + 1 = 1[n + 1] -------eq 2
Combining eq 1 and 2
mn + m + n + 1 = m[n + 1] + 1[n + 1]
= [m + 1][n + 1]
Hence the factors of[mn + m + n + 1] are [m + 1] and [n + 1]
Question 26.
Factories the expressions-
6ab – b2 + 12ac – 2bc
Answer:
In the given expression
Check weather there is any common factors for all terms;
None;
Regrouping the 1st 2 terms we have,
6ab – b2 = b[6a - b] -------eq 1
Regrouping the last 2 terms we have,
12ac – 2bc = 2c[6a - b] -------eq 2
Combining eq 1 and 2
6ab – b2 + 12ac – 2bc = b[6a - b] + 2c[6a - b]
= [6a - b][b + 2c]
Hence the factors of[6ab – b2 + 12ac – 2bc] are [6a - b] and[b + 2c]
Question 27.
Factories the expressions-
p2q – pr2 – pq + r2
Answer:
In the given expression
Check weather there is any common factors for all terms;
None;
Regrouping the 1st 2 terms we have,
p2q – pr2 = p[pq – r2] -------eq 1
Regrouping the last 2 terms we have,
– pq + r2 = -1[pq – r2] -------eq 2
∵ we have to make common parts in both eq 1 and 2
Combining eq 1 and 2
p2q – pr2 – pq + r2 = p[pq – r2] -1[pq – r2]
= [pq – r2][p - 1]
Hence the factors of[p2q – pr2 – pq + r2] are [pq – r2] and [p - 1]
Question 28.
Factories the expressions-
x (y + z) – 5 (y + z)
Answer:
In the given expression
Take out the common in all the terms,
⇒ x (y + z) – 5 (y + z)
⇒ (y + z)(x - 5)
Hence the factors of x (y + z) – 5 (y + z) are (y + z) and (x - 5)
Question 29.
Factories the following
x4 – y4
Answer:
In expression x4 – y4
Both terms are perfect square
⇒ x4 = x2 × x2
⇒ y4 = y2 × y2
∴ x4 – y4 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = x2 and b = y2;
x4 – y4 = (x2 – y2)( x2 + y2),
∴ x2 – y2 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = x and b = y;
x2 – y2 = (x– y)( x+ y),
⇒ x4 – y4 = (x– y)( x+ y), ( x2 + y2)
Hence the factors of x4 – y4 are (x– y),( x+ y) and ( x2 + y2)
Question 30.
Factories the following
a4 – (b + c)4
Answer:
In expression a4 – (b + c)4
Both terms are perfect square
⇒ a4 = a2 × a2
⇒ (b + c)4 = (b + c)2 × (b + c)2
∴ a4 – (b + c)4 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = a2 and b = (b + c)2;
a4 – (b + c)4 = (a2 – (b + c)2)( a2 + (b + c)2),
∴ a2 – (b + c)2 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = a and b = (b + c);
a2 – (b + c)2 = (a– (b + c))( a+ (b + c)),
⇒ a4 – (b + c)4 = (a– (b + c))( a+ (b + c)), ( a2 + (b + c)2)
⇒ a4 – (b + c)4 = (a–b–c)(a + b + c), (a2 + b2 + c2 + 2bc)
Hence the factors of a4 – (b + c)4 are (a–b–c),(a + b + c),( a2 + b2 + c2 + 2bc)
Question 31.
Factories the following
l2 – (m – n)2
Answer:
In the given expression l2 – (m – n)2
Both terms are perfect square
⇒ l2 = l × l
⇒ (m – n)2 = (m – n) × (m – n)
∴ l2 – (m - n)2 Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = a and b = (m - n);
∴ l2 – (m – n)2 = (l + m-n)(l-m + n)
Hence the factors of l2–(m–n)2are (l + m-n)(l-m + n)
Question 32.
Answer:
In the given expression 49x2 –
Both terms are perfect square
⇒ 49x2 = 7x × 7x
⇒ ()2 =
49x2 – Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = 7x and b = ;
∴ (7x)2 –()2 = ( 7x –
) (7x +
)
Hence the factors of 49x2 – are ( 7x –
) and (7x +
)
Question 33.
Factories the following
x4 – 2x2y2 + y4
Answer:
In the given expression
1st and last terms are perfect square
⇒ x4 = x2 × x2
⇒ y4 = y2 × y2
And the middle expression is in form of 2ab
2x2y2 = 2 × x2 × y2
∴ x2 × x2 - 2 × x2 × y2 + y2 × y2
Gives (a-b)2 = a2-2ab + b2
⇒ x4 – 2x2y2 + y4
a = x2 and b = y2;
∴ x4 – 2x2y2 + y4 = (x2 – y2) (x2 + y2)
Hence the factors of x4 – 2x2y2 + y4 are (x2 – y2) and (x2 + y2)
Question 34.
Factories the following
4 (a + b)2 – 9 (a – b)2
Answer:
In the given expression
We know that
(a + b)2 = a2 + 2ab + b2
(a-b)2 = a2-2ab + b2
Hence
4[a2 + 2ab + b2] – 9[a2-2ab + b2]
4a2 + 8ab + 4b2 - 9a2 + 18ab - 9b2
26ab – 5a2 - 5b2
25ab + ab – 5a2 – 5b2
[25ab – 5a2] + [ab – 5b2]
5a[5b – a] – b[5b – a]
[5a – b][5b – a]
Hence the factors 4 (a + b)2 – 9 (a – b)2 are [5a – b] and [5b – a]
Question 35.
Factories the following expressions
a2 + 10a + 24
Answer:
The given expression looks as
x2 + (a + b)x + ab
where a + b = 10; and ab = 24;
factors of 24 their sum
1 × 24 1 + 24 = 25
12 × 2 2 + 12 = 14
6 × 4 6 + 4 = 10
∴ the factors having sum 10 are 6 and 4
a2 + 10a + 24 = a2 + (6 + 4)a + 24
= a2 + 6a + 4a + 24
= a(a + 6) + 4(a + 6)
= (a + 6)(a + 4)
Hence the factors of a2 + 10a + 24 are (a + 6) and (a + 4)
Question 36.
Factories the following expressions
x2 + 9x + 18
Answer:
The given expression looks as
x2 + (a + b)x + ab
where a + b = 9; and ab = 18;
factors of 18 their sum
1 × 18 1 + 18 = 19
9 × 2 2 + 9 = 11
6 × 3 6 + 3 = 9
∴ the factors having sum 9 are 6 and 3
x2 + 9x + 18 = x2 + (6 + 3)x + 18
= x2 + 6x + 3x + 18
= x(x + 6) + 3(x + 6)
= (x + 6)(x + 3)
Hence the factors of x2 + 9x + 18 are (x + 6) and (x + 3)
Question 37.
Factories the following expressions
p2 – 10p + 21
Answer:
The given expression looks as
x2 + (a + b)x + ab
where a + b = -10; and ab = 21;
factors of 21 their sum
-1 × -21 -1-18 = -19
-7 × -3 -7-3 = -10
∴ the factors having sum -10 are -7 and -3
p2 + 9p + 18 = p2 + (-7-3)p + 21
= p2 -7p-3p + 21
= p(p-7) -3(p-7)
= (p-7)(p-3)
Hence the factors of p2 + 9p + 18 are (p-7) and (p-3)
Question 38.
Factories the following expressions
x2 – 4x – 32
Answer:
The given expression looks as
x2 + (a + b)x + ab
where a + b = -4; and ab = -32;
factors of -32 their sum
1 × -32 1-32 = -31
-16 × 2 2 -16 = - 14
-8 × 4 4 -8 = -4
∴ the factors having sum -4 are -8 and 4
x2 – 4x – 32 = x2 + (4 -8)x - 32
= x2 + 4x - 8x - 32
= x(x + 4) -8(x + 4)
= (x + 4)(x-8)
Hence the factors of x2 – 4x – 32 are (x + 4) and (x-8)
Question 39.
The lengths of the sides of a triangle are integrals, and its area is also integer. One side is 21 and the perimeter is 48. Find the shortest side.
Answer:
A = √ s(s-a)(s-b)(s-c)
If the area is an integer
Then [s(s-a)(s-b)(s-c)] should be proper square
If s = Then s =
= 24
Hence ;
A = √ 24(24-a)(24-b)(24-c)
If side of triangle are
a = 21 and b + c = 27
let c be smallest side
then b = 27-c
∴ √ 24(24-21)(24-27 + c)(24-c)
⇒ √ 24 × 3 × (c-3)(24-c)
⇒ √ 2 × 2 × 2 × 3 × 3 × (c-3)(24-c)
⇒ 2 × 3√2(c-3)(24-c)
⇒ 6√2(c-3)(24-c)
∴ the value of [2(c-3)(24-c)] must be a perfect square for area to be a integer
For getting square 2(c-3) should be equal to (24-c)
2(c-3) = (24-c)
2c-6 = 24-c
2c + c = 24 + 6
3c = 10
c = 10; b = 27-c = 27-10 = 17
Hence the size of smallest size is 10.
Question 40.
Find the values of ‘m’ for which x2 + 3xy + x + my –m has two linear factors in x and y, with integer coefficients.
Answer:
For the given 2 degree equation
That must be equal to(ax + by + c)(dx + e)
= ad.x2 + bd.xy + cd.x + ea.x + be.y + ec
= ad.x2 + bd.xy + (cd + ea).x + be.y + ec
x2 + 3xy + x + my–m = ad.x2 + bd.xy + (cd + ea).x + be.y + ec
compare the equation
and take out the coefficient of every term
a.d = 1 ----------1
b.d = 3 ----------2
c.d + e.a = 1 ----------3
b.e = m ----------4
e.c = -m ----------5
⇒ from eq 1; a = d = 1 ∵ all coefficient are integers
After putting result in eq 3; c + e = 1 -------6
After putting result in eq 2; b = 3 --------7
⇒ divide eq 4 and 5
∴ that implies b = -c = -3 ∵ eq 7
Put value of c in eq 6
-3 + e = 1
e = 1 + 3 = 4
Putting value of b and e in eq 4
m = b × e
m = 3 × 4 = 12
Exercise 12.3
Question 1.Carry out the following divisions
48a3 by 6a
Answer:In the given term
Dividend = 48a3 = 2 × 2 × 2 × 2 × 3 × a × a × a
Divisor = 6a = 2 × 3 × a
![](data:image/png;base64,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)
= 2 × 2 × 2 × a × a
= 8a2
Hence dividing 48a3 by 6a gives 8a2
Question 2.Carry out the following divisions
14x3 by 42x2
Answer:In the given term
Dividend = 14x3 = 2 × 7 × x × x × x
Divisor = 42x2 = 2 × 3 × 7 × x × x
= ![](data:image/png;base64,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)
= ![](data:image/png;base64,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)
Hence dividing 14x3 by 42x2 gives ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAcAAAAgCAMAAADpJZJvAAAAAXNSR0IArs4c6QAAAGZQTFRFAAAAAAAAAAA6ADo6ADpmADqQAGa2OgAAOgA6OjoAOma2ZgA6ZpDbZrbbZrb/kDoAkDo6kGY6kLbbkLb/kNv/tmYAttv/25A625Bm27Zm27aQ2////7Zm/7aQ/9uQ/9u2//+2///b6mEI+QAAAAF0Uk5TAEDm2GYAAAAJcEhZcwAADsQAAA7EAZUrDhsAAAAZdEVYdFNvZnR3YXJlAE1pY3Jvc29mdCBPZmZpY2V/7TVxAAAAZ0lEQVQYV42OyQ6AIAwFHyjivqLihvL/P2mtFy8mzGXSdJoU+GHR0jitgF1tc75RNUcltzZTz3gMvoknWJ34UVAXiHgJrD+ZzYSkB3zbY5GGFy5ln131aBQRG1h1jaswcGR6kO9+uAEtUgQwwhfX1AAAAABJRU5ErkJggg==)
Question 3.Carry out the following divisions
72a3b4c5 by 8ab2c3
Answer:In the given term
Dividend = 72a3b4c5 = 2 × 2 × 2 × 3 × 3 × a × a × a × b × b × b × b × c × c × c × c × c
Divisor = 8ab2c3 = 2 × 2 × 2 × a × b × b × c × c × c
= ![](data:image/png;base64,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)
= 3 × 3 × a × a × b × b × c × c
= 9a2b2c2
Hence dividing 72a3b4c5 by 8ab2c3 gives 9a2b2c2
Question 4.Carry out the following divisions
11xy2z3 by 55xyz
Answer:In the given term
Dividend = 11xy2z3 = 11 × x × y × y × z × z × z
Divisor = 55xyz = 5 × 11 × x × y × z
= ![](data:image/png;base64,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)
= ![](data:image/png;base64,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)
= ![](data:image/png;base64,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)
Hence dividing 11xy2z3 by 55xyz gives ![](data:image/png;base64,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)
Question 5.Carry out the following divisions
–54l4m3n2 by 9l2m2n2
Answer:In the given term
Dividend = -54l4m3n2 = (-1) × 2 × 3 × 3 × 3 × l × l × l × l × m × m × m × n × n
Divisor = 9l2m2n2 = 3 × 3 × l × l × m × m × n × n
= ![](data:image/png;base64,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)
= (-1) × 3 × 2 × l × l × m
= -6l2m
Hence dividing –54l4m3n2 by 9l2m2n2 gives -6l2m
Question 6.Divide the given polynomial by the given monomial
(3x2– 2x) ÷ x
Answer:In the given term
Dividend = (3x2– 2x)
Take out the common part in binomial term
= (3 × x × x– 2 × x)
= x(3x-2)
Divisor = x
= ![](data:image/png;base64,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)
= 3x-2
Hence dividing (3x2– 2x) by x gives out 3x-2
Question 7.Divide the given polynomial by the given monomial
(5a3b – 7ab3) ÷ ab
Answer:In the given term
Dividend = (5a3b – 7ab3)
Take out the common part in binomial term
= (5 × a × a × a × b – 7 × a × b × b × b)
= ab(5a2 – 7b2)
Divisor = ab
= ![](data:image/png;base64,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)
= (5a2 – 7b2)
Hence dividing (5a3b – 7ab3) by ab gives out (5a2 – 7b2)
Question 8.Divide the given polynomial by the given monomial
(25x5 – 15x4) ÷ 5x3
Answer:In the given term
Dividend = (25x5 – 15x4)
Take out the common part in binomial term
= (5 × 5 × x × x × x × x × x – 3 × 5 × x × x × x × x)
= (5x – 3)5x4
Divisor = 5x3
= ![](data:image/png;base64,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)
= (5x – 3)x
= 5x2 – 3x
Hence dividing (25x5 – 15x4) by 5x3 gives out 5x2 – 3x
Question 9.Divide the given polynomial by the given monomial
4(l5 – 6l4 + 8l3) ÷ 2l2
Answer:In the given term
Dividend = (4l5 – 6l4 + 8l3)
Take out the common part in binomial term
= (2 × 2 × l × l × l × l × l– 3 × 2 × l × l × l × l + 2 × 2 × 2 × l × l × l )
= (2l2 – 3l + 4)2l3
Divisor = 2l2
= ![](data:image/png;base64,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)
= (2l2 – 3l + 4)l
= (2l3 –2l2 + 4l)
Hence dividing 4(l5 – 6l4 + 8l3) by 2l2 gives out (2l3 –2l2 + 4l)
Question 10.Divide the given polynomial by the given monomial
15 (a3b2c2– a2b3c2 + a2b2c3) ÷ 3abc
Answer:In the given term
Dividend = 15 (a3b2c2– a2b3c2 + a2b2c3)
Take out the common part in binomial term
= 3 × 5(a × a × a × b × b × c × c– a × a × b × b × b × c × c + a × a × b × b × c × c × c )
= 15 a2b2c2(a – b + c)
Divisor = 3abc
= ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAIUAAAApCAMAAAAs2hqbAAAAAXNSR0IArs4c6QAAAJNQTFRFAAAAAAAAAAA6AABmADo6ADpmADqQAGa2OgAAOgBmOjoAOjo6OjpmOjqQOmaQOma2OpC2OpDbZgAAZgA6ZgBmZjoAZjo6ZpDbZrbbZrb/kDoAkDo6kGY6kLa2kNv/tmYAtmY6tmZmtpBmtrZmttv/tv//25A625Bm27Zm27aQ2////7Zm/9uQ/9u2/9vb//+2///bTSHTeAAAAAF0Uk5TAEDm2GYAAAAJcEhZcwAADsQAAA7EAZUrDhsAAAAZdEVYdFNvZnR3YXJlAE1pY3Jvc29mdCBPZmZpY2V/7TVxAAACYUlEQVRYR+1X23LTMBCV3JsphdKUXmJCU6AVbuzU+v+vY3flXV0cj/uAzNCxZvIgaS9nzyrykVLL+H8YsI+lLu4DvMnCYD9Lad31gzKnOx87WRjsZ0GBQc0Z/HQABReC0U/3m5Az2Y88D2A0l77G+sMzWDQaRpxCqfYTmm09Crcgo5+2F0+HeQg8uabIUMK1t+dHhOJqGKh2WH1b+gW27Kf223qkGVFDHbPxMJdu/rprUhR47LD8Bkx+AEJz+qvUaM4L9lHrYi3Thrhir6hh7OkWPYo+guq+IgM4UhS2ulJtuVbdCnqECQz8U6KF7cmTrdZ+H0tkr6hW9uwXPQoXAb2Yxh4FpDxx7X1VgCAgGXm1lWezW8Xd29I09eJe9p62woOnNRHvIzhf4YLKccdwvznXOkGhth4F8hIOF0m8DGUjEzoX3lO48BGGKJQhoG35ZTfBxQEUA6+EC5pyJT68oDDFA/Jwx1w0MK/LO18tGtRB/cTZb3AK2z3worSxp6CQCHwu6J44esYTrj9TQ+xGH3+vEBpn+Zhc5d2NLgKU9B8ZeBGK2NOfTo6A9+9fGuP3xWSC4Q0y6TJqMHp3ToVsL/i6mLJ8y/7Id2TKNfyOTNku+7My4K7VfzxmrXhJtjAwOwNTSjsToJcbkGXBByBR2pmyJmFRiuxXgZxOlPY8KEiDgEIW1S0anZX1TDjsTxBEorpFabOyngeF0QW+l1h1s0ZPtXl2MHQuWHWz0k61eXYUqPFFdXsuxl6KGfB0q3u1h0eSqG5W2qk2z5A7CFnDownOhahuUdqJNs+LYon+7hn4A1kdQmz6BvYVAAAAAElFTkSuQmCC)
= 5abc[a-b + c]
= [5a2bc– 5ab2c + 5abc2]
Hence dividing 15 (a3b2c2– a2b3c2 + a2b2c3) by 3abc gives out [5a2bc– 5ab2c + 5abc2]
Question 11.Divide the given polynomial by the given monomial
(3p3– 9p2q - 6pq2) ÷ (–3p)
Answer:In the given term
Dividend = (3p3– 9p2q - 6pq2)
Take out the common part in binomial term
= (3 × p × p × p– 3 × 3 × p × p × q - 2 × 3 × p × q × q )
= 3 × p(p2– 3pq - 2q2)
Divisor = (–3p)
= ![](data:image/png;base64,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)
= (-1) (p2– 3pq - 2q2)
= (2q2 + 3pq - p2)
Hence dividing (3p3– 9p2q - 6pq2) by (–3p) gives out (2q2 + 3pq - p2)
Question 12.Divide the given polynomial by the given monomial
(
a2b2c2 +
ab2c2) ÷
abc
Answer:In the given term
Dividend = (
a2b2c2 +
ab2c2)
Take out the common part in binomial term
= (
× a × a × b × b × c × c +
× 2 × a × b × b × c × c )
=
× a × b × b × c × c (a + 2)
=
ab �2c2(a + 2)
Divisor =
abc
= ![](data:image/png;base64,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)
= ![](data:image/png;base64,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)
=
bc(a + 2)
Hence dividing (
a2b2c2 +
ab2c2) by
abc gives out
bc(a + 2)
Question 13.Workout the following divisions:
(49x – 63) ÷ 7
Answer:In the given term
Dividend = (49x – 63)
Take out the common part in binomial term
= (7 × 7 × x - 7 × 9)
= 7(7 × x - 9)
= 7(7x - 9)
Divisor = 7
= ![](data:image/png;base64,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)
= (7x - 9)
Hence dividing (49x – 63) by 7 gives out (7x - 9)
Question 14.Workout the following divisions:
12x (8x – 20) ÷ 4(2x – 5)
Answer:In the given term
Dividend = 12x (8x – 20)
Take out the common part in binomial term
= 2 × 2 × 3 × x(2 × 2 × 2 × x - 2 × 2 × 5 )
= 2 × 2 × 2 × 2 × 3 × x(2 × x - 5 )
= 48x (2x - 5 )
Divisor = 4(2x – 5)
![](data:image/png;base64,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)
= 12x
Hence divides 12x (8x – 20) by 4(2x – 5) gives out 12x
Question 15.Workout the following divisions:
11a3b3(7c – 35) ÷ 3a2b2(c – 5)
Answer:In the given term
Dividend = 11a3b3(7c – 35) ÷ 3a2b2(c – 5)
Take out the common part in binomial term
= 11 × a × a × a × b × b × b (7 × c - 5 × 7 )
= 11 × a × a × a × b × b × b × 7 (c - 5 )
= 77a3b3(c - 5)
Divisor = 3a2b2(c – 5)
= ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAE0AAAAnCAMAAAB9uDW+AAAAAXNSR0IArs4c6QAAAJ9QTFRFAAAAAAAAAAA6AABmADo6ADpmADqQAGaQAGa2OgAAOgA6OgBmOjoAOjo6OjqQOmaQOma2OpDbZgAAZgA6ZgBmZjoAZjpmZjqQZmaQZma2ZpDbZrbbZrb/kDoAkDo6kDqQkGY6kLbbkNv/tmYAtmY6tmaQtpA6tpBmttv/tv+2tv//25A625Bm27Zm27aQ2////7Zm/9uQ/9u2//+2///b5pe/0wAAAAF0Uk5TAEDm2GYAAAAJcEhZcwAADsQAAA7EAZUrDhsAAAAZdEVYdFNvZnR3YXJlAE1pY3Jvc29mdCBPZmZpY2V/7TVxAAACF0lEQVRIS91Va1fCMAxtURDfgqIgDJGholO20f7/32YebWnnHuDxcI7kS5dtub1J01wh/p/pmZTtD8s79HbPRt0t1WBi40JvdzSIWF2lIpFdE4sevhwuy8E+R/hdRxKs61ZHpy+v4XvMaIo9kfR+YiUQ35qLHLfTT6nQ07lbzd/qIdURhDKa9SiiaIkpiN0pu+E/aF1D/XtCDfgU4pOFbE2sRwF4JB13QMDXoOXn9BKpuVX1x2LRcRTio6VIjkysjiBQR93UBDIHzBRpGJjMVNquIvPQ4Ft+xrsBEKAZDypJRsTyU9oGnwNqQnxdQiIbblVoQfWoxIwWUsvgeJJN38adVL8Gmap+N9Uvm9NAMswNC1agpiLZfnNdJmJg2jaJmgbJB94biF9Ag4wxXT6F7a02oKwX66ETPLxyK+3FBqINN2v7NPf/J/fgH9n+6R/qjnrq6wtcvMDdNWtSFBgJRhKsWyMJsAOrAtnnhWw9+puqe7y/RhJgjpNbcQ19VUAMPX0WXzCHnOVD2siisVt1DYuqgJE0UFkRRDZKc6TKkmDdUklAyqEqINh6BgPKKAKNaFQTloR3dsslAdFCVcA4eYzjDsxTBMq0VBIMhrQj2VMFRlnhKA4VoRbNPzPg7VQBMrzlwV5QBNEsCXSEvipQh5xShxQUYQtJQLTfqUKThuyoCjWSUNOOQYF9p1ISsFibm1UZf1AfvgF1ykbANJwyhQAAAABJRU5ErkJggg==)
=
ab
Hence dividing 11a3b3(7c – 35) by 3a2b2(c – 5) gives out
ab
Question 16.Workout the following divisions:
54lmn (l + m) (m + n) (n + l) ÷ 81mn (l + m) (n + l)
Answer:In the given term
Dividend = 54lmn (l + m) (m + n) (n + l)
Divisor = 81mn (l + m) (n + l)
= ![](data:image/png;base64,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)
=
l(m + n)
Hence dividing 54lmn (l + m) (m + n) (n + l) by 81mn (l + m)(n + l) gives out
l(m + n)
Question 17.Workout the following divisions:
36 (x + 4) (x2 + 7x + 10) ÷ 9 (x + 4)
Answer:In the given term
Dividend = 36 (x + 4) (x2 + 7x + 10)
Divisor = 9 (x + 4)
= ![](data:image/png;base64,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)
= 4(x2 + 7x + 10)
= (4x2 + 27x + 40)
Hence dividing 36 (x + 4) (x2 + 7x + 10) by 9 (x + 4) gives out
(4x2 + 27x + 40)
Question 18.Workout the following divisions:
a (a + 1) (a + 2) (a + 3) ÷ a (a + 3)
Answer:In the given term
Dividend = a (a + 1) (a + 2) (a + 3)
Divisor = a (a + 3)
= ![](data:image/png;base64,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)
= (a + 1) (a + 2)
Hence dividing a (a + 1) (a + 2) (a + 3) by a (a + 3) gives out
(a + 1) (a + 2)
Question 19.Factorize the expressions and divide them as directed:
(x2 + 7x + 12) ÷ (x + 3)
Answer:In the given term
Dividend = (x2 + 7x + 12)
The given expression looks as
x2 + (a + b)x + ab
where a + b = 7; and ab = 12;
factors of 12 their sum
1 × 12 1 + 12 = 13
6 × 2 2 + 6 = 8
4 × 3 4 + 3 = 7
∴ the factors having sum 7 are 4 and 3
x2 + 7x + 12 = x2 + (4 + 3)x + 12
= x2 + 4x + 3x + 12
= x(x + 4) + 3(x + 4)
= (x + 4)(x + 3)
Divisor = (x + 3)
= ![](data:image/png;base64,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)
= (x + 4)
Hence dividing (x2 + 7x + 12) by (x + 3) gives out (x + 4)
Question 20.Factorize the expressions and divide them as directed:
(x2 – 8x + 12) ÷ (x – 6)
Answer:In the given term
Dividend = (x2 - 8x + 12)
The given expression looks as
x2 + (a + b)x + ab
where a + b = -8; and ab = 12;
factors of 12 their sum
-1 × -12 -1-12 = -13
-6 × -2 -2-6 = -8
-4 × -3 -4-3 = -7
∴ the factors having sum 7 are 4 and 3
x2 - 8x + 12 = x2 + (-6-2)x + 12
= x2 - 6x - 2x + 12
= x(x - 6) -2(x - 6)
= (x - 6)(x - 2)
Divisor = (x - 6)
= ![](data:image/png;base64,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)
= (x – 2)
Hence dividing (x2 – 8x + 12) by (x – 6) gives out (x – 2)
Question 21.Factorize the expressions and divide them as directed:
(p2 + 5p + 4) ÷ (p + 1)
Answer:In the given term
Dividend = (p2 + 5p + 4)
The given expression looks as
x2 + (a + b)x + ab
where a + b = 5; and ab = 4;
factors of 4 their sum
1 × 4 1 + 4 = 5
2 × 2 2 + 2 = 4
∴ the factors having sum 5 are 4 and 1
(p2 + 5p + 4) = p2 + (4 + 1)p + 4
= p2 + 4p + p + 4
= p(p + 4) + 1(p + 4)
= (p + 1)(p + 4)
Divisor = (p + 1)
= ![](data:image/png;base64,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)
= (p + 4)
Hence dividing (p2 + 5p + 4) by (p + 1) gives out (p + 4)
Question 22.Factorize the expressions and divide them as directed:
15ab (a2–7a + 10) ÷ 3b (a – 2)
Answer:In the given term
Dividend = 15ab (a2–7a + 10)
The given expression (a2–7a + 10) looks as
x2 + (a + b) x + ab
where a + b = -7; and ab = 10;
factors of 10 their sum
-1 × -10 -1-10 = -11
-2 × -5 -2-5 = -7
∴ the factors having sum -7 are -2 and -5
(a2–7a + 10) = a2 + (-2-5)a + 10
= a2–5a – 2a + 10
= a(a – 5) – 2(a – 5)
= (a – 5)(a – 2)
Divisor = 3b (a – 2)
= ![](data:image/png;base64,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)
= 5a(a – 5)
Hence dividing 15ab (a2–7a + 10) by 3b (a – 2) gives out 5a(a – 5)
Question 23.Factorize the expressions and divide them as directed:
15lm (2p2–2q2) ÷ 3l (p + q)
Answer:In the given term
Dividend = 15lm (2p2–2q2)
In given expression (2p2–2q2)
Take out the common factor in binomial term
⇒ (2 × p × p – 2 × q × q)
→ 2(p2 – q2)
Both terms are perfect square
⇒ p2 = p × p
⇒ q2 = q × q
∴ (p2 – q2) Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = p and b = q;
p2 – q2 = (p + q)(p – q)
Hence the factors of p2 – q2 are (p + q) and (p – q)
Divisor = 3l (p + q)
= ![](data:image/png;base64,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)
= ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAJAAAAAlCAMAAACan5BMAAAAAXNSR0IArs4c6QAAAKJQTFRFAAAAAAAAAAA6AABmADo6ADqQAGaQAGa2OgAAOgA6OgBmOjoAOjo6OjqQOmZmOmaQOma2OpDbZgAAZgA6ZjoAZjo6ZjqQZmaQZma2ZpDbZra2ZrbbZrb/kDoAkDo6kGY6kLbbkLb/kNv/tmYAtmY6tmZmtpA6tpBmtpxmttv/tv//25A625Bm27Zm27aQ2//b2////7Zm/9uQ/9u2//+2///b2BAl0gAAAAF0Uk5TAEDm2GYAAAAJcEhZcwAADsQAAA7EAZUrDhsAAAAZdEVYdFNvZnR3YXJlAE1pY3Jvc29mdCBPZmZpY2V/7TVxAAACVklEQVRYR+1Xa3ObMBCUaG1o+kjiOkkfdpMG51W5Lsbw//9a73RCCAmoruVDmrFmNIO5095qT8BaiOP4JwV2V09D67efikFoJdMwNoJFyQ1ivdGr62uZfPZg1PnIdsr3IaPtWwLJQ0KjWFSGELcflnq1SovyzW2HQF9JJ8GU2KzgXnWJWPW3G/EzARBFhJxQiFWvcWFnGES9ulqsRL0+F3s5e8iSm3uZfBcY38s5XLvSHZZSvsMl5ckPhKsuVjDtVvSuDCEnFGK1hHxEIvQRcFBpBTvMcaa0Qr26pV2bUa/TAqbWg+5WF1+AkxmH669abvppQz1YllCA6BGaFULhbAjBNchXLaQeq1YBi1hm9qzl8rVLSDQhQ4iwdEMMHCptNTWInZYRmZCQVSgk1FFI7DIoOKSQJYQCN2fIR6zzeQHxfEaHuiG0mf8yLXvSbWxGtTgrDm7L8PzgYTF9L5HQRkPqo0WhpmUOliXkISrUDiSvljIBtctMpnu4gVMfeown7vOwz+TpvXOo75AsPWVim+kHABfjTycUYrWH2ke0m/cv9GOEigVDtyTizdIuHMainDhEdQbq9xHCLv/hLeVvYhCLEiMRd1ePsuczgC0Z+3T0aj6ARbl/hTjY2pcdoPfT8xkvW+3j7o4KxCrQWuZonxwLzczzLXPE14z5geER8i0zxyfzKjGyXcvM8cmMEqxU1zKzfDKrSnxyxzKzfHJ8DWamY5l9QoA06JOZVSLT8a+SY5kZPjmyADvNs8yT+GQ2iZEF0/jkKRlN45OnZHT0yVOq+Z9i/QbnXV1SFDtQeQAAAABJRU5ErkJggg==)
= 10m(p – q)
Hence dividing 15lm (2p2–2q2) by 3l (p + q) gives out 10m(p – q)
Question 24.Factorize the expressions and divide them as directed:
26z3(32z2–18) ÷ 13z2(4z – 3)
Answer:In the given term
Dividend = 26z3(32z2–18)
Take out the common factor in binomial term
⇒ 2 × 13 × z × z × z (2 × 2 × 2 × 2 × 2 × z × z – 2 × 3 × 3)
⇒ 2 × 2 × 13 × z × z × z (2 × 2 × 2 × 2 × z × z – 3 × 3)
⇒ 52z3(16z2 – 9)
In given expression (16z2 – 9)
Both terms are perfect square
⇒ 16z2 = 4z × 4z
⇒ 9 = 3 × 3
∴ (16z2 – 9) Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = 4z and b = 3;
(16z2 – 9) = (4z + 3)(4z – 3)
Hence the factors of (16z2 – 9)are (4z + 3) and (4z – 3)
Divisor = 13z2(4z – 3)
= ![](data:image/png;base64,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)
= ![](data:image/png;base64,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)
= 4z(4z + 3)
Hence dividing 26z3(32z2–18) by 13z2(4z – 3) gives out 4z(4z + 3)
Carry out the following divisions
48a3 by 6a
Answer:
In the given term
Dividend = 48a3 = 2 × 2 × 2 × 2 × 3 × a × a × a
Divisor = 6a = 2 × 3 × a
= 2 × 2 × 2 × a × a
= 8a2
Hence dividing 48a3 by 6a gives 8a2
Question 2.
Carry out the following divisions
14x3 by 42x2
Answer:
In the given term
Dividend = 14x3 = 2 × 7 × x × x × x
Divisor = 42x2 = 2 × 3 × 7 × x × x
=
=
Hence dividing 14x3 by 42x2 gives
Question 3.
Carry out the following divisions
72a3b4c5 by 8ab2c3
Answer:
In the given term
Dividend = 72a3b4c5 = 2 × 2 × 2 × 3 × 3 × a × a × a × b × b × b × b × c × c × c × c × c
Divisor = 8ab2c3 = 2 × 2 × 2 × a × b × b × c × c × c
=
= 3 × 3 × a × a × b × b × c × c
= 9a2b2c2
Hence dividing 72a3b4c5 by 8ab2c3 gives 9a2b2c2
Question 4.
Carry out the following divisions
11xy2z3 by 55xyz
Answer:
In the given term
Dividend = 11xy2z3 = 11 × x × y × y × z × z × z
Divisor = 55xyz = 5 × 11 × x × y × z
=
=
=
Hence dividing 11xy2z3 by 55xyz gives
Question 5.
Carry out the following divisions
–54l4m3n2 by 9l2m2n2
Answer:
In the given term
Dividend = -54l4m3n2 = (-1) × 2 × 3 × 3 × 3 × l × l × l × l × m × m × m × n × n
Divisor = 9l2m2n2 = 3 × 3 × l × l × m × m × n × n
=
= (-1) × 3 × 2 × l × l × m
= -6l2m
Hence dividing –54l4m3n2 by 9l2m2n2 gives -6l2m
Question 6.
Divide the given polynomial by the given monomial
(3x2– 2x) ÷ x
Answer:
In the given term
Dividend = (3x2– 2x)
Take out the common part in binomial term
= (3 × x × x– 2 × x)
= x(3x-2)
Divisor = x
=
= 3x-2
Hence dividing (3x2– 2x) by x gives out 3x-2
Question 7.
Divide the given polynomial by the given monomial
(5a3b – 7ab3) ÷ ab
Answer:
In the given term
Dividend = (5a3b – 7ab3)
Take out the common part in binomial term
= (5 × a × a × a × b – 7 × a × b × b × b)
= ab(5a2 – 7b2)
Divisor = ab
=
= (5a2 – 7b2)
Hence dividing (5a3b – 7ab3) by ab gives out (5a2 – 7b2)
Question 8.
Divide the given polynomial by the given monomial
(25x5 – 15x4) ÷ 5x3
Answer:
In the given term
Dividend = (25x5 – 15x4)
Take out the common part in binomial term
= (5 × 5 × x × x × x × x × x – 3 × 5 × x × x × x × x)
= (5x – 3)5x4
Divisor = 5x3
=
= (5x – 3)x
= 5x2 – 3x
Hence dividing (25x5 – 15x4) by 5x3 gives out 5x2 – 3x
Question 9.
Divide the given polynomial by the given monomial
4(l5 – 6l4 + 8l3) ÷ 2l2
Answer:
In the given term
Dividend = (4l5 – 6l4 + 8l3)
Take out the common part in binomial term
= (2 × 2 × l × l × l × l × l– 3 × 2 × l × l × l × l + 2 × 2 × 2 × l × l × l )
= (2l2 – 3l + 4)2l3
Divisor = 2l2
=
= (2l2 – 3l + 4)l
= (2l3 –2l2 + 4l)
Hence dividing 4(l5 – 6l4 + 8l3) by 2l2 gives out (2l3 –2l2 + 4l)
Question 10.
Divide the given polynomial by the given monomial
15 (a3b2c2– a2b3c2 + a2b2c3) ÷ 3abc
Answer:
In the given term
Dividend = 15 (a3b2c2– a2b3c2 + a2b2c3)
Take out the common part in binomial term
= 3 × 5(a × a × a × b × b × c × c– a × a × b × b × b × c × c + a × a × b × b × c × c × c )
= 15 a2b2c2(a – b + c)
Divisor = 3abc
=
= 5abc[a-b + c]
= [5a2bc– 5ab2c + 5abc2]
Hence dividing 15 (a3b2c2– a2b3c2 + a2b2c3) by 3abc gives out [5a2bc– 5ab2c + 5abc2]
Question 11.
Divide the given polynomial by the given monomial
(3p3– 9p2q - 6pq2) ÷ (–3p)
Answer:
In the given term
Dividend = (3p3– 9p2q - 6pq2)
Take out the common part in binomial term
= (3 × p × p × p– 3 × 3 × p × p × q - 2 × 3 × p × q × q )
= 3 × p(p2– 3pq - 2q2)
Divisor = (–3p)
=
= (-1) (p2– 3pq - 2q2)
= (2q2 + 3pq - p2)
Hence dividing (3p3– 9p2q - 6pq2) by (–3p) gives out (2q2 + 3pq - p2)
Question 12.
Divide the given polynomial by the given monomial
( a2b2c2 +
ab2c2) ÷
abc
Answer:
In the given term
Dividend = ( a2b2c2 +
ab2c2)
Take out the common part in binomial term
= ( × a × a × b × b × c × c +
× 2 × a × b × b × c × c )
= × a × b × b × c × c (a + 2)
= ab �2c2(a + 2)
Divisor = abc
=
=
= bc(a + 2)
Hence dividing ( a2b2c2 +
ab2c2) by
abc gives out
bc(a + 2)
Question 13.
Workout the following divisions:
(49x – 63) ÷ 7
Answer:
In the given term
Dividend = (49x – 63)
Take out the common part in binomial term
= (7 × 7 × x - 7 × 9)
= 7(7 × x - 9)
= 7(7x - 9)
Divisor = 7
=
= (7x - 9)
Hence dividing (49x – 63) by 7 gives out (7x - 9)
Question 14.
Workout the following divisions:
12x (8x – 20) ÷ 4(2x – 5)
Answer:
In the given term
Dividend = 12x (8x – 20)
Take out the common part in binomial term
= 2 × 2 × 3 × x(2 × 2 × 2 × x - 2 × 2 × 5 )
= 2 × 2 × 2 × 2 × 3 × x(2 × x - 5 )
= 48x (2x - 5 )
Divisor = 4(2x – 5)
= 12x
Hence divides 12x (8x – 20) by 4(2x – 5) gives out 12x
Question 15.
Workout the following divisions:
11a3b3(7c – 35) ÷ 3a2b2(c – 5)
Answer:
In the given term
Dividend = 11a3b3(7c – 35) ÷ 3a2b2(c – 5)
Take out the common part in binomial term
= 11 × a × a × a × b × b × b (7 × c - 5 × 7 )
= 11 × a × a × a × b × b × b × 7 (c - 5 )
= 77a3b3(c - 5)
Divisor = 3a2b2(c – 5)
=
=
ab
Hence dividing 11a3b3(7c – 35) by 3a2b2(c – 5) gives out ab
Question 16.
Workout the following divisions:
54lmn (l + m) (m + n) (n + l) ÷ 81mn (l + m) (n + l)
Answer:
In the given term
Dividend = 54lmn (l + m) (m + n) (n + l)
Divisor = 81mn (l + m) (n + l)
=
=
l(m + n)
Hence dividing 54lmn (l + m) (m + n) (n + l) by 81mn (l + m)(n + l) gives out l(m + n)
Question 17.
Workout the following divisions:
36 (x + 4) (x2 + 7x + 10) ÷ 9 (x + 4)
Answer:
In the given term
Dividend = 36 (x + 4) (x2 + 7x + 10)
Divisor = 9 (x + 4)
=
= 4(x2 + 7x + 10)
= (4x2 + 27x + 40)
Hence dividing 36 (x + 4) (x2 + 7x + 10) by 9 (x + 4) gives out
(4x2 + 27x + 40)
Question 18.
Workout the following divisions:
a (a + 1) (a + 2) (a + 3) ÷ a (a + 3)
Answer:
In the given term
Dividend = a (a + 1) (a + 2) (a + 3)
Divisor = a (a + 3)
=
= (a + 1) (a + 2)
Hence dividing a (a + 1) (a + 2) (a + 3) by a (a + 3) gives out
(a + 1) (a + 2)
Question 19.
Factorize the expressions and divide them as directed:
(x2 + 7x + 12) ÷ (x + 3)
Answer:
In the given term
Dividend = (x2 + 7x + 12)
The given expression looks as
x2 + (a + b)x + ab
where a + b = 7; and ab = 12;
factors of 12 their sum
1 × 12 1 + 12 = 13
6 × 2 2 + 6 = 8
4 × 3 4 + 3 = 7
∴ the factors having sum 7 are 4 and 3
x2 + 7x + 12 = x2 + (4 + 3)x + 12
= x2 + 4x + 3x + 12
= x(x + 4) + 3(x + 4)
= (x + 4)(x + 3)
Divisor = (x + 3)
=
= (x + 4)
Hence dividing (x2 + 7x + 12) by (x + 3) gives out (x + 4)
Question 20.
Factorize the expressions and divide them as directed:
(x2 – 8x + 12) ÷ (x – 6)
Answer:
In the given term
Dividend = (x2 - 8x + 12)
The given expression looks as
x2 + (a + b)x + ab
where a + b = -8; and ab = 12;
factors of 12 their sum
-1 × -12 -1-12 = -13
-6 × -2 -2-6 = -8
-4 × -3 -4-3 = -7
∴ the factors having sum 7 are 4 and 3
x2 - 8x + 12 = x2 + (-6-2)x + 12
= x2 - 6x - 2x + 12
= x(x - 6) -2(x - 6)
= (x - 6)(x - 2)
Divisor = (x - 6)
=
= (x – 2)
Hence dividing (x2 – 8x + 12) by (x – 6) gives out (x – 2)
Question 21.
Factorize the expressions and divide them as directed:
(p2 + 5p + 4) ÷ (p + 1)
Answer:
In the given term
Dividend = (p2 + 5p + 4)
The given expression looks as
x2 + (a + b)x + ab
where a + b = 5; and ab = 4;
factors of 4 their sum
1 × 4 1 + 4 = 5
2 × 2 2 + 2 = 4
∴ the factors having sum 5 are 4 and 1
(p2 + 5p + 4) = p2 + (4 + 1)p + 4
= p2 + 4p + p + 4
= p(p + 4) + 1(p + 4)
= (p + 1)(p + 4)
Divisor = (p + 1)
=
= (p + 4)
Hence dividing (p2 + 5p + 4) by (p + 1) gives out (p + 4)
Question 22.
Factorize the expressions and divide them as directed:
15ab (a2–7a + 10) ÷ 3b (a – 2)
Answer:
In the given term
Dividend = 15ab (a2–7a + 10)
The given expression (a2–7a + 10) looks as
x2 + (a + b) x + ab
where a + b = -7; and ab = 10;
factors of 10 their sum
-1 × -10 -1-10 = -11
-2 × -5 -2-5 = -7
∴ the factors having sum -7 are -2 and -5
(a2–7a + 10) = a2 + (-2-5)a + 10
= a2–5a – 2a + 10
= a(a – 5) – 2(a – 5)
= (a – 5)(a – 2)
Divisor = 3b (a – 2)
=
= 5a(a – 5)
Hence dividing 15ab (a2–7a + 10) by 3b (a – 2) gives out 5a(a – 5)
Question 23.
Factorize the expressions and divide them as directed:
15lm (2p2–2q2) ÷ 3l (p + q)
Answer:
In the given term
Dividend = 15lm (2p2–2q2)
In given expression (2p2–2q2)
Take out the common factor in binomial term
⇒ (2 × p × p – 2 × q × q)
→ 2(p2 – q2)
Both terms are perfect square
⇒ p2 = p × p
⇒ q2 = q × q
∴ (p2 – q2) Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = p and b = q;
p2 – q2 = (p + q)(p – q)
Hence the factors of p2 – q2 are (p + q) and (p – q)
Divisor = 3l (p + q)
=
=
= 10m(p – q)
Hence dividing 15lm (2p2–2q2) by 3l (p + q) gives out 10m(p – q)
Question 24.
Factorize the expressions and divide them as directed:
26z3(32z2–18) ÷ 13z2(4z – 3)
Answer:
In the given term
Dividend = 26z3(32z2–18)
Take out the common factor in binomial term
⇒ 2 × 13 × z × z × z (2 × 2 × 2 × 2 × 2 × z × z – 2 × 3 × 3)
⇒ 2 × 2 × 13 × z × z × z (2 × 2 × 2 × 2 × z × z – 3 × 3)
⇒ 52z3(16z2 – 9)
In given expression (16z2 – 9)
Both terms are perfect square
⇒ 16z2 = 4z × 4z
⇒ 9 = 3 × 3
∴ (16z2 – 9) Seems to be in identity a2-b2 = (a + b)(a-b)
Where a = 4z and b = 3;
(16z2 – 9) = (4z + 3)(4z – 3)
Hence the factors of (16z2 – 9)are (4z + 3) and (4z – 3)
Divisor = 13z2(4z – 3)
=
=
= 4z(4z + 3)
Hence dividing 26z3(32z2–18) by 13z2(4z – 3) gives out 4z(4z + 3)
Exercise 12.4
Question 1.Find the errors and correct the following mathematical sentences
3(x – 9) = 3x – 9
Answer:If LHS is
3(x – 9)
Then RHS would be
⇒ 3(x – 9)
= 3 × x – 3 × 9
= 3x – 27
The error is 27 instead of 9
Hence 3(x – 9) = 3x – 27
Question 2.Find the errors and correct the following mathematical sentences
x(3x + 2) = 3x2 + 2
Answer:If LHS is
x(3x + 2)
Then RHS would be
⇒ x(3x + 2)
= 3 × x × x – 2 × x
= 3x2 – 2x
The error is 2x instead of 2
Hence x(3x + 2) = 3x2 + 2x
Question 3.Find the errors and correct the following mathematical sentences
2x + 3x = 5x2
Answer:If LHS is
2x + 3x
Then RHS would be
⇒ 2x + 3x
= x(2 + 3)
= 5x
The error is 5x instead of 5x2
Hence 2x + 3x = 5x
Question 4.Find the errors and correct the following mathematical sentences
2x + x + 3x = 5x
Answer:If LHS is
2x + x + 3x = 5x
Then RHS would be
⇒ 2x + x + 3x
= x(2 + 1 + 3)
= 6x
The error is 6x instead of 5x
Hence 2x + x + 3x = 6x
Question 5.Find the errors and correct the following mathematical sentences
4p + 3p + 2p + p – 9p = 0
Answer:If LHS is
4p + 3p + 2p + p – 9p
Then RHS would be
⇒ 4p + 3p + 2p + p – 9p
= p(4 + 3 + 2 + 1–9)
= p(10–9)
= p
The error is p instead of 0
Hence 4p + 3p + 2p + p–9p = p
Question 6.Find the errors and correct the following mathematical sentences
3x + 2y = 6xy
Answer:If RHS is
6xy
Then LHS would be
⇒ 6xy
= 2 × 3 × x × y
= 3 × x × 2 × y
= 3x × 2y
The error is sign of multiplication instead of sign of addition
Hence 3x × 2y = 6xy
Question 7.Find the errors and correct the following mathematical sentences
(3x)2 + 4x + 7 = 3x2 + 4x + 7
Answer:If LHS is
(3x)2 + 4x + 7
Then RHS would be
⇒ (3x)2 + 4x + 7
= 32 × x2 + 4x + 7
= 9x2 + 4x + 7
The error is 9x2 instead of 3x2
Hence (3x)2 + 4x + 7 = 9x2 + 4x + 7
Question 8.Find the errors and correct the following mathematical sentences
(2x)2 + 5x = 4x + 5x = 9x
Answer:If LHS is
(2x)2 + 5x
Then RHS would be
⇒ (2x)2 + 5x
= 22 × x2 + 5x
= 4x2 + 5x
The error is 4x2 instead of 4x
Hence (2x)2 + 5x = 4x2 + 5x
Question 9.Find the errors and correct the following mathematical sentences
(2a + 3)2 = 2a2 + 6a + 9
Answer:If LHS is
(2a + 3)2
Then RHS would be
⇒ (2a + 3)2
= (2a)2 + 32 + 2 × 2a × 3
= 4a2 + 9 + 12a
= 4a2 + 12a + 9
The error is 4a2 instead of 2a2 and 12a instead of 6a
Hence = (2a + 3)2 = 4a2 + 9 + 12a
Question 10.Find the errors and correct the following mathematical sentences
Substitute x = – 3 in
(a) x2 + 7x + 12 = (–3)2 + 7 (–3) + 12 = 9 + 4 + 12 = 25
Answer:If LHS is
x2 + 7x + 12
Then RHS would be
⇒ x2 + 7x + 12
Putting x = (-3)
= (–3)2 + 7 (–3) + 12
= 9 + (-21) + 12
= 21-21
= 0
The error is (-21) instead of 4 and end result 0 instead of 25
Hence putting x = (-3) in x2 + 7x + 12 results to 0
Question 11.Find the errors and correct the following mathematical sentences
Substitute x = – 3 in
(b) x2– 5x + 6 = (–3)2 –5 (–3) + 6 = 9 – 15 + 6 = 0
Answer:If LHS is
x2– 5x + 6
Then RHS would be
⇒ x2– 5x + 6
Putting x = (-3)
= (–3)2 –5 (–3) + 6
= 9 + 15 + 6
= 30
The error is + 15 instead of (-15) and end results to 30 instead of 0
Hence putting x = (-3) in x2– 5x + 6 results to 30
Question 12.Find the errors and correct the following mathematical sentences
Substitute x = – 3 in
(c) x2 + 5x = (–3)2 + 5 (–3) + 6 = – 9 – 15 = –24
Answer:If LHS is
x2 + 5x
Then RHS would be
⇒ x2 + 5x
Putting x = (-3)
= (–3)2 + 5 (–3)
= 9 + (-15)
= -6
The error is ( + 9) instead of (-9) and end results to (-6) instead of (-24)
Hence putting x = (-3) in x2 + 5x results to (-6)
Question 13.Find the errors and correct the following mathematical sentences
(x – 4)2 = x2 – 16
Answer:If LHS is
(x – 4)2
Then RHS would be
⇒ (x – 4)2
= (x)2 + 42 – 2 × x × 4
= x2 + 16 – 8x
The error is x2 + 16 – 8x instead of x2 – 16
Hence (x – 4)2 = x2 + 13 – 8x
Question 14.Find the errors and correct the following mathematical sentences
(x + 7)2 = x2 + 49
Answer:If LHS is
(x + 7)2
Then RHS would be
⇒ (x + 7)2
= (x)2 + 72 + 2 × x × 7
= x2 + 49 + 14
The error is x2 + 14x + 49 instead of x2 + 49
Hence (x + 7)2 = x2 + 14x + 49
Question 15.Find the errors and correct the following mathematical sentences
(3a + 4b) (a – b) = 3a2 – 4a2
Answer:For getting in the equation
(a2 – b2 ) = (a + b)(a-b)
RHS would be
3a2 – 4b2
Then LHS would be
⇒ 3a2 – 4b2
= (3a – 4b)(3a + 4b)
The error is (a – b) instead of (3a – 4b)
3a2 – 4b2 instead of 3a2 – 4a2
Hence 3a2 – 4b2 = (3a – 4b)(3a + 4b)
Question 16.Find the errors and correct the following mathematical sentences
(x + 4) (x + 2) = x2 + 8
Answer:If LHS is
(x + 4) (x + 2)
Then RHS would be
⇒ (x + 4) (x + 2)
= x2 + 4 × x + 2 × x + 2 × 4
= x2 + 4x + 2x + 8
= x2 + 6x + 8
The error is x2 + 6x + 8 instead of x2 + 8
Hence (x + 4) (x + 2) = x2 + 6x + 8
Question 17.Find the errors and correct the following mathematical sentences
(x – 4) (x – 2) = x2 – 8
Answer:If LHS is
(x – 4) (x – 2)
Then RHS would be
⇒ (x – 4) (x – 2)
= x2 – 4 × x – 2 × x + (-2) × (-4)
= x2 – 4x – 2x + 8
= x2 – 6x + 8
The error is x2 – 6x + 8 instead of x2 – 8
Hence (x – 4) (x – 2) = x2 – 6x + 8
Question 18.Find the errors and correct the following mathematical sentences
5x3 ÷ 5x3 = 0
Answer:If LHS is
5x3 ÷ 5x3
Then RHS would be
⇒ 5x3 ÷ 5x3
= ![](data:image/png;base64,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)
= 1
The error is1 instead of 0
Hence 5x3 ÷ 5x3 = 1
Question 19.Find the errors and correct the following mathematical sentences
2x3 + 1 ÷ 2x3 = 1
Answer:If LHS is
(2x3 + 1) ÷ 2x3
Then RHS would be
⇒ (2x3 + 1) ÷ 2x3
= ![](data:image/png;base64,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)
= ![](data:image/png;base64,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)
= ![](data:image/png;base64,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)
The error is
instead of 1
Hence (2x3 + 1) ÷ 2x3 = ![](data:image/png;base64,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)
Question 20.Find the errors and correct the following mathematical sentences
3x + 2 ÷ 3x = ![](data:image/png;base64,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)
Answer:If LHS is
(3x + 2) ÷ 3x
Then RHS would be
⇒ (3x + 2) ÷ 3x
= ![](data:image/png;base64,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)
= ![](data:image/png;base64,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)
= ![](data:image/png;base64,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)
The error is
instead of![](data:image/png;base64,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)
Hence (3x + 2 )÷ 3x = ![](data:image/png;base64,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)
Question 21.Find the errors and correct the following mathematical sentences
3x + 5 ÷ 3 = 5
Answer:If LHS is
For the complete and perfect division
There must be 3x instead of x
(3x + 5)÷3x
Then RHS would be
⇒ (3x + 5)÷3x
= ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAACQAAAAiCAMAAAAAh4u3AAAAAXNSR0IArs4c6QAAAGxQTFRFAAAAAAAAAAA6ADo6ADpmADqQAGa2OgAAOgA6OjoAOma2ZgAAZgA6ZpDbZrbbZrb/kDoAkDo6kGY6kLbbkLb/kNv/tmYAtmY6ttv/25A625Bm27Zm27aQ2////7Zm/7aQ/9uQ/9u2//+2///bGXy4RwAAAAF0Uk5TAEDm2GYAAAAJcEhZcwAADsQAAA7EAZUrDhsAAAAZdEVYdFNvZnR3YXJlAE1pY3Jvc29mdCBPZmZpY2V/7TVxAAAAxElEQVQ4T9WS3Q6CMAyFuyFO8A/UoRsik73/O9qNkeA2uMDEhF4t7VlP23wAfw6ZEXqc9BSEEAq6vEJN+bdKF2eXEMMD1I5DzShXLG1sMSJ6X05YeKWNyHvNWIR2OWYqkhgRiKQfrdtjAWMwUsymn8wkZObMxp1sW+gOHJQRtTddbB6enS5NFRswewLJtroi/uD6jheww6wz+sPNxzo3WzT1PO4OkBjuMbcA94goxD0U+bhPbBbg7uuiuAfNYrgvuuVvnz64FAzjI9pzSAAAAABJRU5ErkJggg==)
= ![](data:image/png;base64,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)
= ![](data:image/png;base64,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)
The error is
instead of 5 and 3x instead of x
Hence = (3x + 5)÷3x = ![](data:image/png;base64,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)
Question 22.Find the errors and correct the following mathematical sentences
= x + 1
Answer:If LHS is
![](data:image/png;base64,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)
Then RHS would be
⇒ ![](data:image/png;base64,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)
=
x + ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAcAAAAiCAMAAACk7TNkAAAAAXNSR0IArs4c6QAAAEVQTFRFAAAAAAAAAAA6ADo6OgAAOgA6OjoAZgA6ZpDbZrbbZrb/kDo6kGY6kLbbkNv/tmYA25Bm27Zm27aQ/9uQ/9u2//+2///beZmaKAAAAAF0Uk5TAEDm2GYAAAAJcEhZcwAADsQAAA7EAZUrDhsAAAAZdEVYdFNvZnR3YXJlAE1pY3Jvc29mdCBPZmZpY2V/7TVxAAAARUlEQVQoU2NgwAUE2BiZOBgYxLh5GQSZ+MCqhFnBtAgPJ4jiZ2QG0wwMQixcDKLsfAzCQJpBgAWsj1jACAHEKkeoo437AIHhAtLt5ojxAAAAAElFTkSuQmCC)
=
x + 1
The error is
x + 1 instead of x + 1
Hence
x + 1
Find the errors and correct the following mathematical sentences
3(x – 9) = 3x – 9
Answer:
If LHS is
3(x – 9)
Then RHS would be
⇒ 3(x – 9)
= 3 × x – 3 × 9
= 3x – 27
The error is 27 instead of 9
Hence 3(x – 9) = 3x – 27
Question 2.
Find the errors and correct the following mathematical sentences
x(3x + 2) = 3x2 + 2
Answer:
If LHS is
x(3x + 2)
Then RHS would be
⇒ x(3x + 2)
= 3 × x × x – 2 × x
= 3x2 – 2x
The error is 2x instead of 2
Hence x(3x + 2) = 3x2 + 2x
Question 3.
Find the errors and correct the following mathematical sentences
2x + 3x = 5x2
Answer:
If LHS is
2x + 3x
Then RHS would be
⇒ 2x + 3x
= x(2 + 3)
= 5x
The error is 5x instead of 5x2
Hence 2x + 3x = 5x
Question 4.
Find the errors and correct the following mathematical sentences
2x + x + 3x = 5x
Answer:
If LHS is
2x + x + 3x = 5x
Then RHS would be
⇒ 2x + x + 3x
= x(2 + 1 + 3)
= 6x
The error is 6x instead of 5x
Hence 2x + x + 3x = 6x
Question 5.
Find the errors and correct the following mathematical sentences
4p + 3p + 2p + p – 9p = 0
Answer:
If LHS is
4p + 3p + 2p + p – 9p
Then RHS would be
⇒ 4p + 3p + 2p + p – 9p
= p(4 + 3 + 2 + 1–9)
= p(10–9)
= p
The error is p instead of 0
Hence 4p + 3p + 2p + p–9p = p
Question 6.
Find the errors and correct the following mathematical sentences
3x + 2y = 6xy
Answer:
If RHS is
6xy
Then LHS would be
⇒ 6xy
= 2 × 3 × x × y
= 3 × x × 2 × y
= 3x × 2y
The error is sign of multiplication instead of sign of addition
Hence 3x × 2y = 6xy
Question 7.
Find the errors and correct the following mathematical sentences
(3x)2 + 4x + 7 = 3x2 + 4x + 7
Answer:
If LHS is
(3x)2 + 4x + 7
Then RHS would be
⇒ (3x)2 + 4x + 7
= 32 × x2 + 4x + 7
= 9x2 + 4x + 7
The error is 9x2 instead of 3x2
Hence (3x)2 + 4x + 7 = 9x2 + 4x + 7
Question 8.
Find the errors and correct the following mathematical sentences
(2x)2 + 5x = 4x + 5x = 9x
Answer:
If LHS is
(2x)2 + 5x
Then RHS would be
⇒ (2x)2 + 5x
= 22 × x2 + 5x
= 4x2 + 5x
The error is 4x2 instead of 4x
Hence (2x)2 + 5x = 4x2 + 5x
Question 9.
Find the errors and correct the following mathematical sentences
(2a + 3)2 = 2a2 + 6a + 9
Answer:
If LHS is
(2a + 3)2
Then RHS would be
⇒ (2a + 3)2
= (2a)2 + 32 + 2 × 2a × 3
= 4a2 + 9 + 12a
= 4a2 + 12a + 9
The error is 4a2 instead of 2a2 and 12a instead of 6a
Hence = (2a + 3)2 = 4a2 + 9 + 12a
Question 10.
Find the errors and correct the following mathematical sentences
Substitute x = – 3 in
(a) x2 + 7x + 12 = (–3)2 + 7 (–3) + 12 = 9 + 4 + 12 = 25
Answer:
If LHS is
x2 + 7x + 12
Then RHS would be
⇒ x2 + 7x + 12
Putting x = (-3)
= (–3)2 + 7 (–3) + 12
= 9 + (-21) + 12
= 21-21
= 0
The error is (-21) instead of 4 and end result 0 instead of 25
Hence putting x = (-3) in x2 + 7x + 12 results to 0
Question 11.
Find the errors and correct the following mathematical sentences
Substitute x = – 3 in
(b) x2– 5x + 6 = (–3)2 –5 (–3) + 6 = 9 – 15 + 6 = 0
Answer:
If LHS is
x2– 5x + 6
Then RHS would be
⇒ x2– 5x + 6
Putting x = (-3)
= (–3)2 –5 (–3) + 6
= 9 + 15 + 6
= 30
The error is + 15 instead of (-15) and end results to 30 instead of 0
Hence putting x = (-3) in x2– 5x + 6 results to 30
Question 12.
Find the errors and correct the following mathematical sentences
Substitute x = – 3 in
(c) x2 + 5x = (–3)2 + 5 (–3) + 6 = – 9 – 15 = –24
Answer:
If LHS is
x2 + 5x
Then RHS would be
⇒ x2 + 5x
Putting x = (-3)
= (–3)2 + 5 (–3)
= 9 + (-15)
= -6
The error is ( + 9) instead of (-9) and end results to (-6) instead of (-24)
Hence putting x = (-3) in x2 + 5x results to (-6)
Question 13.
Find the errors and correct the following mathematical sentences
(x – 4)2 = x2 – 16
Answer:
If LHS is
(x – 4)2
Then RHS would be
⇒ (x – 4)2
= (x)2 + 42 – 2 × x × 4
= x2 + 16 – 8x
The error is x2 + 16 – 8x instead of x2 – 16
Hence (x – 4)2 = x2 + 13 – 8x
Question 14.
Find the errors and correct the following mathematical sentences
(x + 7)2 = x2 + 49
Answer:
If LHS is
(x + 7)2
Then RHS would be
⇒ (x + 7)2
= (x)2 + 72 + 2 × x × 7
= x2 + 49 + 14
The error is x2 + 14x + 49 instead of x2 + 49
Hence (x + 7)2 = x2 + 14x + 49
Question 15.
Find the errors and correct the following mathematical sentences
(3a + 4b) (a – b) = 3a2 – 4a2
Answer:
For getting in the equation
(a2 – b2 ) = (a + b)(a-b)
RHS would be
3a2 – 4b2
Then LHS would be
⇒ 3a2 – 4b2
= (3a – 4b)(3a + 4b)
The error is (a – b) instead of (3a – 4b)
3a2 – 4b2 instead of 3a2 – 4a2
Hence 3a2 – 4b2 = (3a – 4b)(3a + 4b)
Question 16.
Find the errors and correct the following mathematical sentences
(x + 4) (x + 2) = x2 + 8
Answer:
If LHS is
(x + 4) (x + 2)
Then RHS would be
⇒ (x + 4) (x + 2)
= x2 + 4 × x + 2 × x + 2 × 4
= x2 + 4x + 2x + 8
= x2 + 6x + 8
The error is x2 + 6x + 8 instead of x2 + 8
Hence (x + 4) (x + 2) = x2 + 6x + 8
Question 17.
Find the errors and correct the following mathematical sentences
(x – 4) (x – 2) = x2 – 8
Answer:
If LHS is
(x – 4) (x – 2)
Then RHS would be
⇒ (x – 4) (x – 2)
= x2 – 4 × x – 2 × x + (-2) × (-4)
= x2 – 4x – 2x + 8
= x2 – 6x + 8
The error is x2 – 6x + 8 instead of x2 – 8
Hence (x – 4) (x – 2) = x2 – 6x + 8
Question 18.
Find the errors and correct the following mathematical sentences
5x3 ÷ 5x3 = 0
Answer:
If LHS is
5x3 ÷ 5x3
Then RHS would be
⇒ 5x3 ÷ 5x3
=
= 1
The error is1 instead of 0
Hence 5x3 ÷ 5x3 = 1
Question 19.
Find the errors and correct the following mathematical sentences
2x3 + 1 ÷ 2x3 = 1
Answer:
If LHS is
(2x3 + 1) ÷ 2x3
Then RHS would be
⇒ (2x3 + 1) ÷ 2x3
=
=
=
The error is instead of 1
Hence (2x3 + 1) ÷ 2x3 =
Question 20.
Find the errors and correct the following mathematical sentences
3x + 2 ÷ 3x =
Answer:
If LHS is
(3x + 2) ÷ 3x
Then RHS would be
⇒ (3x + 2) ÷ 3x
=
=
=
The error is instead of
Hence (3x + 2 )÷ 3x =
Question 21.
Find the errors and correct the following mathematical sentences
3x + 5 ÷ 3 = 5
Answer:
If LHS is
For the complete and perfect division
There must be 3x instead of x
(3x + 5)÷3x
Then RHS would be
⇒ (3x + 5)÷3x
=
=
=
The error is instead of 5 and 3x instead of x
Hence = (3x + 5)÷3x =
Question 22.
Find the errors and correct the following mathematical sentences = x + 1
Answer:
If LHS is
Then RHS would be
⇒
= x +
= x + 1
The error is x + 1 instead of x + 1
Hence x + 1