Factorisation Class 8th Mathematics AP Board Solution

Class 8^th 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…

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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 1^st 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 1^st 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 1^st 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 1^st 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 1^st 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]

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Exercise 12.2

Question 1.

Factories the following expression-

a² + 10a + 25

Answer:

In the given expression

1^st 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

1^st 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

1^st 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

1^st 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

1^st 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

1^st 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

1^st 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

1^st 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

1^st 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 1^st 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 1^st 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 1^st 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 1^st 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 1^st 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

1^st 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)

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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