# GSEB Solutions Class 8 Maths Chapter 14 Factorization Ex 14.3

Gujarat BoardĀ GSEB Textbook Solutions Class 8 Maths Chapter 14 Factorization Ex 14.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 8 Maths Chapter 14 Factorization Ex 14.3

Question 1.
Carry out the following divisions.

1. 28x4 + 56x
2. -36y3 + 9y2
3. 66pq2r3 + 11qr2
4. 34x3y3z3 + 51xy2z3
5. 12a8b8 + (-6a6b4)

Solution:
1. We have 28x4 Ć· 56x = $$\frac{28 x^{4}}{56 x}$$

ā“ 28x4 Ć· 56x = $$\frac{1}{2}$$x3

2. We have
-36y3 = (-1) Ć 2 Ć 2 Ć 3 Ć 3 Ć y Ć y Ć y and 9y2 = 3 Ć 3 Ć y Ć y

3. We have
66pq2r3 = 2 Ć 3 Ć 11 Ć p Ć q Ć q Ć r Ć r Ć r
and 11qr2 = 11 Ć q Ć r Ć r

4. We have

5. We have

Question 2.
Divide the given polynomial by the given monomial?

1. (5x2 – 6x) Ć· 3x
2. (3y8 – 4y6 + 5y4) Ć· y4
3. 8(x3y2z2 + x2y3z2 + x2y2z3) Ć· 4x3y2z2
4. (x3 + 2x2 + 3x) Ć· 2x
5. (p3q6 – p6q3) Ć· p3q3

Solution:

Question 3.
Work out the following divisions:

1. (10x – 25) Ć· 5
2. (10x – 25) Ć· (2x – 5)
3. 10y(6y + 21) Ć· 5(2y + 7)
4. 9x2y2(3z- 24) Ć· 27xy(z – 8)
5. 96abc(3a – 12)(5b – 30) Ć· 144(1 – 4)(b – 6)

Solution:
1. āµ 10x – 25 = 5(2x – 5)
ā“ $$\frac{10x-25}{5}$$ = $$\frac{5(2x-5)}{5}$$ = (2x – 5)
Thus, (10x – 25) Ć· 5 = 2x – 5

2. āµ 10x – 25 = 5(2x – 5)
ā“ $$\frac{10x-25}{2x-5}$$ = $$\frac{5(2x-5)}{(2x-5}$$ = 5
Thus, (10x – 25) Ć· (2x – 5) = 5

3. āµ 6y + 21 = 3(2y + 7)
ā“ $$\frac{10y(6y+21)}{5(2y+7)}$$ = $$\frac{10yĆ3Ć(2y+7)}{5(2y+7)}$$
= 2 Ć y Ć 3 = 6y
ā“ 10y(6y + 21) Ć· 5(2y + 7) = 6y

4. āµ 3z – 24 = 3(z – 8)

ā“ 9x2y2(3z – 24) Ć· 27xy(z – 8) = xy

5. āµ 3a – 12 = 3(a – 4)
5b – 30 = 5(b – 6)
96 = 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 3 and 144 = 2 Ć 2 Ć 2 Ć 2 Ć 3 Ć 3

Thus, 96abc (3a – 12)(5b – 30) Ć· 144(a – 4)(b – 6) = 10abc

Question 4.
Divide as directed.

1. 5(2x + 1)(3x + 5) Ć· (2x + 1)
2. 26xy(x + 5)(y – 4) Ć· 13x(y – 4)
3. 52pqr(p + q)(q + r)(r + p) Ć· 104pq(q + r)(r + p)
4. 20(y + 4)(y2 + 5y + 3) Ć· 5(y + 4)
5. x(x + 1)(x + 2) (x + 3) Ć· x(x + 1)

Solution:
1. We have 5(2x + 1)(3x +5) Ć· (2x + 1)
= $$\frac{5(2x+1)(3x+5)}{(2x+1)}$$
= $$\frac{5Ć(3x+5)}{1}$$ = 5(3x + 5)
ā“ 5(2x + 1)(3x + 5) Ć· (2x + 1)
= 5(3x + 5)

2. We have 26xy(x + 5)(y – 4) Ć· 13x(y – 4)
= $$\frac{26xy(x + 5)(y – 4)}{13x(y – 4)}$$ = $$\frac{2Ćy(x + 5)}{1}$$
ā“ 26xy(x + 5)(y – 4) Ć· 13x(y – 4)
= 2y(x + 5)

3. We have

ā“ 52pqr(p + q)(q + r)(r + p) Ć· 104pq(q + r)
(r + p) = $$\frac{1}{2}$$r(p + q)

4. āµ 20 = 2 Ć 2 Ć 5

= 2 Ć 2 Ć (y2 + 5y + 3) = 4(y2 + 5y + 3)

5. x(x + 1)(x + 2)(x + 3) Ć· x(x + 1)
We have $$\frac{x(x+1)(x+2)(x+3)}{x(x+1)}$$
= $$\frac{(x+2)(x+3)}{1}$$
ā“ x(x + 1)(x + 2)(x + 3) Ć· x(x + 1)
= (x + 2)(x + 3)

Question 5.
Factorise the expressions and divide them as directed.

1. (y2 + 7y + 10) Ć· (y + 5)
2. (m2 – 14m – 32) Ć· (m + 2)
3. (5p2 – 25p + 20) Ć· (p – 1)
4. 4yz(z2 + 6z – 16) Ć· 2y(z + 8)
5. 5pq(p2 – q2) Ć· 2p(p + q)
6. 12xy(9x2 – 16y2) Ć· 4xy(3x + 4y)
7. 39y3(50y2 – 98) Ć· 26y2(5y + 7)

Solution:
1. āµ y2 + 7y + 10 = y2 + 2y + 5y + 10
[Splitting 7y in 2y + 5y such that 2y Ć 5y = 10y2]
= y(y + 2) + 5(y + 2) = (y + 2)(y + 5)
ā“ $$\frac{y^{2}+7 y+10}{(y+5)}$$ = $$\frac{(y+2)(y+5)}{(y+5)}$$ = y + 2
ā“ (y2 + 7y + 10) Ć· (y + 5) = y + 2

2. āµ m2 – 14m – 32
= (m2 – 16m + 2m – 32) [āµ -14 = -16 + 2]
= m(m – 16) + 2(m – 16) [16 Ć 2 = 32]
= (m – 16)(m + 2)
ā“ $$\frac{m^{2}-14m-32}{m+2}$$ = $$\frac{(m-16)(m+2)}{(m+2)}$$ = $$\frac{m-16}{1}$$
Thus, (m2 – 14m – 32) Ć· (m + 2) = m – 16

3. āµ 5p2 – 25p + 20 = 5(p2 – 5p + 4)
[āµ 1 Ć 4 = 4]
= 5[p(p – 1) – 4(p – 1)][-1 + (-4) = -5]
= 5[(p – 1)(p – 4)]
ā“ $$\frac{5 p^{2}-25 p+20}{p-1}$$ = 5$$\frac{(p-1)(p-4)}{(p-1)}$$
= $$\frac{5Ć(p-4)}{1}$$
Thus, (5p2 – 25p + 20) Ć· (p – 1) = 5(p – 4)

4. āµ z2 + 6z – 16 = z2 + 8z – 2z – 16
ā“ 8 – 2 = 6
= z(z + 8) – 2(z + 8)8 Ć 2 = 16
= (z + 8)(z – 2)

Thus, 4yz (z2 + 6z – 16) Ć· 2y(z + 8)
= 2z(z – 2)

5. āµ p2 – q2 = (p – q)(p + q)
[Using a2 – b2 = (a + b)(a – b)]

Thus, 5pq(p2 – q2) Ć· 2p(p + q) = $$\frac{5}{2}$$q(p – q)

6. āµ 9x2 – 16y2 = (3x)2 – (4y)2
= (3x – 4y)(3x + 4y)

ā“ 12xy(9x2 – 16y2) Ć· 4xy(3x + 4y)
= 3(3x – 4y)

7. āµ 50y2 – 98 = 2(25y2 – 49)
= 2[(5y)2 – (7)2] = 2[(5y – 7)(5y + 7)]

Thus, 39y3(50y2 – 98) Ć· 26y2(5y + 7)
= 3y(5y – 7)