GSEB Solutions Class 8 Maths Chapter 14 Factorization Ex 14.3

Gujarat Board 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 ÷ 56x = \(\frac{28 x^{4}}{56 x}\)

∴ 28x⁴ ÷ 56x = \(\frac{1}{2}\)x³

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

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

4. We have

5. We have

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

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

Solution:

Question 3.
Work out the following divisions:

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

Solution:
1. ∵ 10x – 25 = 5(2x – 5)
∴ \(\frac{10x-25}{5}\) = \(\frac{5(2x-5)}{5}\) = (2x – 5)
Thus, (10x – 25) ÷ 5 = 2x – 5

2. ∵ 10x – 25 = 5(2x – 5)
∴ \(\frac{10x-25}{2x-5}\) = \(\frac{5(2x-5)}{(2x-5}\) = 5
Thus, (10x – 25) ÷ (2x – 5) = 5

3. ∵ 6y + 21 = 3(2y + 7)
∴ \(\frac{10y(6y+21)}{5(2y+7)}\) = \(\frac{10y×3×(2y+7)}{5(2y+7)}\)
= 2 × y × 3 = 6y
∴ 10y(6y + 21) ÷ 5(2y + 7) = 6y

4. ∵ 3z – 24 = 3(z – 8)

∴ 9x²y²(3z – 24) ÷ 27xy(z – 8) = xy

5. ∵ 3a – 12 = 3(a – 4)
5b – 30 = 5(b – 6)
96 = 2 × 2 × 2 × 2 × 2 × 3 and 144 = 2 × 2 × 2 × 2 × 3 × 3

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

Question 4.
Divide as directed.

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

Solution:
1. We have 5(2x + 1)(3x +5) ÷ (2x + 1)
= \(\frac{5(2x+1)(3x+5)}{(2x+1)}\)
= \(\frac{5×(3x+5)}{1}\) = 5(3x + 5)
∴ 5(2x + 1)(3x + 5) ÷ (2x + 1)
= 5(3x + 5)

2. We have 26xy(x + 5)(y – 4) ÷ 13x(y – 4)
= \(\frac{26xy(x + 5)(y – 4)}{13x(y – 4)}\) = \(\frac{2×y(x + 5)}{1}\)
∴ 26xy(x + 5)(y – 4) ÷ 13x(y – 4)
= 2y(x + 5)

3. We have

∴ 52pqr(p + q)(q + r)(r + p) ÷ 104pq(q + r)
(r + p) = \(\frac{1}{2}\)r(p + q)

4. ∵ 20 = 2 × 2 × 5

= 2 × 2 × (y² + 5y + 3) = 4(y² + 5y + 3)

5. x(x + 1)(x + 2)(x + 3) ÷ x(x + 1)
We have \(\frac{x(x+1)(x+2)(x+3)}{x(x+1)}\)
= \(\frac{(x+2)(x+3)}{1}\)
∴ x(x + 1)(x + 2)(x + 3) ÷ x(x + 1)
= (x + 2)(x + 3)

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

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

Solution:
1. ∵ y² + 7y + 10 = y² + 2y + 5y + 10
[Splitting 7y in 2y + 5y such that 2y × 5y = 10y²]
= y(y + 2) + 5(y + 2) = (y + 2)(y + 5)
∴ \(\frac{y^{2}+7 y+10}{(y+5)}\) = \(\frac{(y+2)(y+5)}{(y+5)}\) = y + 2
∴ (y² + 7y + 10) ÷ (y + 5) = y + 2

2. ∵ m² – 14m – 32
= (m² – 16m + 2m – 32) [∵ -14 = -16 + 2]
= m(m – 16) + 2(m – 16) [16 × 2 = 32]
= (m – 16)(m + 2)
∴ \(\frac{m^{2}-14m-32}{m+2}\) = \(\frac{(m-16)(m+2)}{(m+2)}\) = \(\frac{m-16}{1}\)
Thus, (m² – 14m – 32) ÷ (m + 2) = m – 16

3. ∵ 5p² – 25p + 20 = 5(p² – 5p + 4)
[∵ 1 × 4 = 4]
= 5[p(p – 1) – 4(p – 1)][-1 + (-4) = -5]
= 5[(p – 1)(p – 4)]
∴ \(\frac{5 p^{2}-25 p+20}{p-1}\) = 5\(\frac{(p-1)(p-4)}{(p-1)}\)
= \(\frac{5×(p-4)}{1}\)
Thus, (5p² – 25p + 20) ÷ (p – 1) = 5(p – 4)

4. ∵ z² + 6z – 16 = z² + 8z – 2z – 16
∴ 8 – 2 = 6
= z(z + 8) – 2(z + 8)8 × 2 = 16
= (z + 8)(z – 2)

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

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

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

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

∴ 12xy(9x² – 16y²) ÷ 4xy(3x + 4y)
= 3(3x – 4y)

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

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