Factorisation Class 8th Mathematics AP Board Solution

Class 8th Mathematics AP Board Solution
Exercise 12.1
  1. 8x, 24 Find the common factors of the given terms in each.
  2. 3a, 21ab Find the common factors of the given terms in each.
  3. 7xy, 35x^2 y^3 Find the common factors of the given terms in each.…
  4. 4m^2 , 6m^2 , 8m^3 Find the common factors of the given terms in each.…
  5. 15p, 20qr, 25rp Find the common factors of the given terms in each.…
  6. 4x^2 , 6xy, 8y^2 x Find the common factors of the given terms in each.…
  7. 12x^2 y, 18xy^2 Find the common factors of the given terms in each.…
  8. 5x^2 - 25xy Factorise the following expressions
  9. 9a^2 - 6ax Factorise the following expressions
  10. 7p^2 + 49pq Factorise the following expressions
  11. 36a^2 b - 60 a^2 bc Factorise the following expressions
  12. 3a^2 bc + 6ab^2 c + 9abc^2 Factorise the following expressions
  13. 4p^2 + 5pq - 6pq^2 Factorise the following expressions
  14. ut + at^2 Factorise the following expressions
  15. 3ax - 6xy + 8by - 4ab Factorise the following:
  16. x^3 + 2x^2 + 5x + 10 Factorise the following:
  17. m^2 - mn + 4m - 4n Factorise the following:
  18. a^3 - a^2 b^2 - ab + b^3 Factorise the following:
  19. p^2 q - pr^2 - pq + r^2 Factorise the following:
Exercise 12.2
  1. a^2 + 10a + 25 Factories the following expression-
  2. l^2 - 16l + 64 Factories the following expression-
  3. 36x^2 + 96xy + 64y^2 Factories the following expression-
  4. 25x^2 + 9y^2 - 30xy Factories the following expression-
  5. 25m^2 - 40mn + 16n^2 Factories the following expression-
  6. 81x^2 - 198 xy + 121y^2 Factories the following expression-
  7. (x + y)^2 - 4xy (Hint: first expand (x + y)^2 Factories the following…
  8. l^4 + 4l^2 m^2 + 4m^4 Factories the following expression-
  9. x^2 - 36 Factories the following
  10. 49x^2 - 25y^2 Factories the following
  11. m^2 - 121 Factories the following
  12. 81 - 64x^2 Factories the following
  13. x^2 y^2 - 64 Factories the following
  14. 6x^2 - 54 Factories the following
  15. x^2 - 81 Factories the following
  16. 2x - 32x^5 Factories the following
  17. 81x^4 - 121x^2 Factories the following
  18. (p^2 - 2pq + q^2) - r^2 Factories the following
  19. (x + y)^2 - (x - y)^2 Factories the following
  20. lx^2 + mx Factories the expressions-
  21. 7y^2 + 35z^2 Factories the expressions-
  22. 3x^4 + 6x^3 y + 9x^2 z Factories the expressions-
  23. x^2 - ax - bx + ab Factories the expressions-
  24. 3ax - 6ay - 8by + 4bx Factories the expressions-
  25. mn + m + n + 1 Factories the expressions-
  26. 6ab - b^2 + 12ac - 2bc Factories the expressions-
  27. p^2 q - pr^2 - pq + r^2 Factories the expressions-
  28. x (y + z) - 5 (y + z) Factories the expressions-
  29. x^4 - y^4 Factories the following
  30. a^4 - (b + c)^4 Factories the following
  31. l^2 - (m - n)^2 Factories the following
  32. 49x^2 - 16/25 Factories the following
  33. x^4 - 2x^2 y^2 + y^4 Factories the following
  34. 4 (a + b)^2 - 9 (a - b)^2 Factories the following
  35. a^2 + 10a + 24 Factories the following expressions
  36. x^2 + 9x + 18 Factories the following expressions
  37. p^2 - 10p + 21 Factories the following expressions
  38. x^2 - 4x - 32 Factories the following expressions
  39. The lengths of the sides of a triangle are integrals, and its area is also…
  40. Find the values of ‘m’ for which x^2 + 3xy + x + my -m has two linear factors…
Exercise 12.3
  1. 48a^3 by 6a Carry out the following divisions
  2. 14x^3 by 42x^2 Carry out the following divisions
  3. 72a^3 b^4 c^5 by 8ab^2 c^3 Carry out the following divisions
  4. 11xy^2 z^3 by 55xyz Carry out the following divisions
  5. -54l^4 m^3 n^2 by 9l^2 m^2 n^2 Carry out the following divisions
  6. (3x^2 - 2x) ÷ x Divide the given polynomial by the given monomial…
  7. (5a^3 b - 7ab^3) ÷ ab Divide the given polynomial by the given monomial…
  8. (25x^5 - 15x^4) ÷ 5x^3 Divide the given polynomial by the given monomial…
  9. 4(l^5 - 6l^4 + 8l^3) ÷ 2l^2 Divide the given polynomial by the given monomial…
  10. 15 (a^3 b^2 c^2 - a^2 b^3 c^2 + a^2 b^2 c^3) ÷ 3abc Divide the given…
  11. (3p^3 - 9p^2 q - 6pq^2) ÷ (-3p) Divide the given polynomial by the given…
  12. (2/3 a^2 b^2 c^2 + 4/3 ab^2 c^2) ÷ 1/2 abc Divide the given polynomial by the…
  13. (49x - 63) ÷ 7 Workout the following divisions:
  14. 12x (8x - 20) ÷ 4(2x - 5) Workout the following divisions:
  15. 11a^3 b^3 (7c - 35) ÷ 3a^2 b^2 (c - 5) Workout the following divisions:…
  16. 54lmn (l + m) (m + n) (n + l) ÷ 81mn (l + m) (n + l) Workout the following…
  17. 36 (x + 4) (x^2 + 7x + 10) ÷ 9 (x + 4) Workout the following divisions:…
  18. a (a + 1) (a + 2) (a + 3) ÷ a (a + 3) Workout the following divisions:…
  19. (x^2 + 7x + 12) ÷ (x + 3) Factorize the expressions and divide them as…
  20. (x^2 - 8x + 12) ÷ (x - 6) Factorize the expressions and divide them as…
  21. (p^2 + 5p + 4) ÷ (p + 1) Factorize the expressions and divide them as…
  22. 15ab (a^2 -7a + 10) ÷ 3b (a - 2) Factorize the expressions and divide them as…
  23. 15lm (2p^2 -2q^2) ÷ 3l (p + q) Factorize the expressions and divide them as…
  24. 26z^3 (32z^2 -18) ÷ 13z^2 (4z - 3) Factorize the expressions and divide them…
Exercise 12.4
  1. 3(x - 9) = 3x - 9 Find the errors and correct the following mathematical…
  2. x(3x + 2) = 3x^2 + 2 Find the errors and correct the following mathematical…
  3. 2x + 3x = 5x^2 Find the errors and correct the following mathematical…
  4. 2x + x + 3x = 5x Find the errors and correct the following mathematical…
  5. 4p + 3p + 2p + p - 9p = 0 Find the errors and correct the following…
  6. 3x + 2y = 6xy Find the errors and correct the following mathematical sentences…
  7. (3x)^2 + 4x + 7 = 3x^2 + 4x + 7 Find the errors and correct the following…
  8. (2x)^2 + 5x = 4x + 5x = 9x Find the errors and correct the following…
  9. (2a + 3)^2 = 2a^2 + 6a + 9 Find the errors and correct the following…
  10. Substitute x = - 3 in (a) x^2 + 7x + 12 = (-3)^2 + 7 (-3) + 12 = 9 + 4 + 12 =…
  11. Substitute x = - 3 in (b) x^2 - 5x + 6 = (-3)^2 -5 (-3) + 6 = 9 - 15 + 6 = 0…
  12. Substitute x = - 3 in (c) x^2 + 5x = (-3)^2 + 5 (-3) + 6 = - 9 - 15 = -24 Find…
  13. (x - 4)^2 = x^2 - 16 Find the errors and correct the following mathematical…
  14. (x + 7)^2 = x^2 + 49 Find the errors and correct the following mathematical…
  15. (3a + 4b) (a - b) = 3a^2 - 4a^2 Find the errors and correct the following…
  16. (x + 4) (x + 2) = x^2 + 8 Find the errors and correct the following…
  17. (x - 4) (x - 2) = x^2 - 8 Find the errors and correct the following…
  18. 5x^3 ÷ 5x^3 = 0 Find the errors and correct the following mathematical…
  19. 2x^3 + 1 ÷ 2x^3 = 1 Find the errors and correct the following mathematical…
  20. 3x + 2 ÷ 3x = 2/3x Find the errors and correct the following mathematical…
  21. 3x + 5 ÷ 3 = 5 Find the errors and correct the following mathematical…
  22. 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]




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.

Factories the following

49x2 –


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



= 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



= 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


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