PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.3

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 14 Factorization Ex 14.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.3

1. Carry out the following divisions:

Question (i)
28x⁴ ÷ 56x
Solution:

Question (ii)
– 36y³ ÷ 9y²
Solution:

Question (iii)
66pq²r³ ÷ 11qr²
Solution:

Question (iv)
34x³y³z³ ÷ 51 xy²z³
Solution:

Question (v)
12a⁸b⁸ ÷ (- 6a⁶b⁴)
Solution:

2. Divide the given polynomial by the given monomial:

Question (i)
(5x² – 6x) ÷ 3x
Solution:

Question (ii)
(3y⁸ – 4y⁶ + 5y⁴) ÷ y⁴
Solution:

Question (iii)
8 (x³y²z² + x²y³z² ÷ 4 x²y²z²)
Solution:

Question (iv)
(x³ + 2x² + 3x) ÷ 2x
Solution:

Question (v)
(P³ q⁶ – p⁶ q³) ÷ p³ q³
Solution:

3. Work out the following divisions:

Question (i)
(10x – 25) ÷ 5
Solution:
= \(\frac{10 x-25}{5}\)
= \(\frac{5(2 x-5)}{5}\)
= 2x – 5

Question (ii)
(10x-25) ÷ (2x – 5)
Solution:
= \(\frac{10 x-25}{2 x-5}\)
= \(\frac{5(2 x-5)}{(2 x-5)}\)
= 5

Question (iii)
10y (6y + 21) ÷ 5 (2y + 7)
Solution:
= \(\frac{10 y(6 y+21)}{5(2 y+7)}\)
= \(\frac{2 \times 5 \times y \times 3 \times(2 y+7)}{5(2 y+7)}\)
= 2 × y × 3
= 6y

Question (iv)
9x²y²(3z – 24) ÷ 27xy (z – 8)
Solution:

Question (v)
96 abc (3a – 12) (5b – 30) ÷ 144 (a – 4)(b – 6)
Solution:
= \(\frac{96 a b c(3 a-12)(5 b-30)}{144(a-4)(b-6)}\)
= \(\frac{2 \times 2 \times 2 \times 2 \times 2 \times 3 \times a \times b \times c \times 3 \times(a-4) \times 5 \times(b-6)}{2 \times 2 \times 2 \times 2 \times 3 \times 3 \times(a-4) \times(b-6)}\)
= 2 × 5 × a × b × c
= 10 abc

4. Divide as directed:

Question (i)
5 (2x + 1) (3x + 5) ÷ (2x + 1)
Solution:
= \(\frac{5(2 x+1)(3 x+5)}{(2 x+1)}\)
= \(\frac{5 \times(3 x+5)}{1}\)
= 5(3x + 5)

Question (ii)
26xy (x + 5) (y – 4) ÷ 13x (y – 4)
Solution:
= \(\frac{26 x y(x+5)(y-4)}{13 x(y-4)}\)
= \(\frac{2 \times 13 \times x \times y(x+5)(y-4)}{13 x(y-4)}\)
= 2y(x + 5)

Question (iii)
52pqr (p + q) (q + r) (r + p) ÷ 104pq (q + r) (r + p)
Solution:
= \(\frac{52 p q r(p+q)(q+r)(r+p)}{104 p q(q+r)(r+p)}\)
= \(\frac{52 \times p \times q \times r \times(p+q)(q+r)(r+p)}{2 \times 52 \times p \times q \times(q+r)(r+p)}\)
= \(\frac{r \times(p+q)}{2}\)
= \(\frac {1}{2}\)r (p + q)

Question (iv)
20 (y + 4) (y² + 5y + 3) ÷ 5 (y + 4)
Solution:
= \(\frac{20(y+4)\left(y^{2}+5 y+3\right)}{5(y+4)}\)
= \(\frac{2 \times 2 \times 5 \times(y+4)\left(y^{2}+5 y+3\right)}{5(y+4)}\)
= 2 × 2 × (y² + 5y + 3)
= 4(y² + 5y + 3)

Question (v)
x (x + 1) (x + 2) (x + 3) ÷ x (x + 1)
Solution:
= \(\frac{x(x+1)(x+2)(x+3)}{x(x+1)}\)
= \(\frac{(x+2)(x+3)}{1}\)
= (x + 2)(x + 3)

5. Factorise the expressions and divide them as directed:

Question (i)
(y² + 7y + 10) ÷ (y + 5)
Solution:
First we factorise
y² + 7y + 10
= y² + 5y + 2y + 10
= y (y + 5) + 2 (y + 5)
= (y + 5) (y + 2)
∴ (y² + 7y + 10) ÷ (y + 5)
= \(\frac{(y+5)(y+2)}{(y+5)}\)
= y + 2

Question (ii)
(m² – 14m – 32) ÷ (m + 2)
Solution:
First we factorise
m² – 14m – 32
= m² – 16m + 2m – 32
= m (m – 16) + 2 (m – 16)
= (m – 16) (m + 2)
∴ (m² – 14m – 32) ÷ (m + 2)
= \(\frac{(m-16)(m+2)}{m+2}\)
= m – 16

Question (iii)
(5p² – 25p + 20) ÷ (p – 1)
Solution:
First we factorise
5p² – 25p + 20
= 5 (p² – 5p + 4)
= 5 (p² – 4p – p + 4)
= 5 [p (p – 4) – 1 (p – 4)]
= 5 (p – 4) (p – 1)
∴ (5p² – 25p + 20) ÷ (p – 1)
= \(\frac{5(p-4)(p-1)}{p-1}\)
= 5 (p – 4)

Question (iv)
4yz (z² + 6z – 16) ÷ 2y (z + 8)
Solution:
= \(\frac{4 y z\left(z^{2}+6 z-16\right)}{2 y(z+8)}\)
= \(\frac{2 z\left(z^{2}+6 z-16\right)}{z+8}\)
Now, factorise
z² + 6z – 16
= z² + 8z – 2z – 16
= z (z + 8) – 2 (z + 8)
= (z + 8) (z – 2)
∴ 4yz (z² + 6z – 16) ÷ 2y (z + 8)
= \(\frac{2 z(z+8)(z-2)}{z+8}\)
= 2z (z – 2)

Question (v)
5pq (p² – q²) ÷ 2p(p + q)
Solution:

Question (vi)
12xy (9x² – 16y²) ÷ 4xy (3x + 4y)
Solution:
= \(\frac{12 x y\left(9 x^{2}-16 y^{2}\right)}{4 x y(3 x+4 y)}\)
= \(\frac{12 x y(3 x-4 y)(3 x+4 y)}{4 x y(3 x+4 y)}\)
= 3 (3x – 4y)

Question (vii)
39y³ (50y² – 98) ÷ 26y² (5y + 7)
Solution: