PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.1

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 14 Factorization Ex 14.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.1

1. Find the common factors of the given terms:

Question (i)
12x, 36
Solution:
12x = 2 × 2 × 3 × x
36 = 2 × 2 × 3 × 3
∴ Common factors of 12x and 36 = 2 × 2 × 3
= 12

Question (ii)
2y, 22xy
Solution:
2y = 2 xy
22xy = 2 × 11 × x × y
∴ Common factors of 2y and 22xy
= 2 × y = 2y

Question (iii)
14pq, 28p²q²
Solution:
14pq = 2 × 7 × p × q
28p²q² = 2 × 2 × 7 × p × p × q × q
∴ Common factors of 14pq and 28p²q²
= 2 × 7 × p × q = 14pq

Question (iv)
2x, 3x², 4
Solution:
2x = 2 × x
3x² = 3 × x × x
4 = 2 × 2
∴ Common factors of 2x, 3x² and 4 = 1 [Note: 1 is a factor of each term.]

Question (v)
6abc, 24ab², 12a²b
Solution:
6abc = 2 × 3 × a × b × c
24ab² = 2 × 2 × 2 × 3 × a × b × b
12a²b = 2 × 2 × 3 × a × a × b
∴ Common factors of 6abc, 24ab² and 12a²b
= 2 × 3 × a × b
= 6ab

Question (vi)
16x³, – 4x², 32x
Solution:
16x³ = 2 × 2 × 2 × 2 × x × x × x
– 4x² = (-1) × 2 × 2 × x × x
32x = 2 × 2 × 2 × 2 × 2 × x
∴ Common factors of 16x³, – 4x² and 32x = 2 × 2 × x = 4x

Question (vii)
10pq, 20qr, 30rp
Solution:
10pq = 2 × 5 × p × q
20qr = 2 × 2 × 5 × q × r
30rp = 2 × 3 × 5 × r × p
∴ Common factors of 10pq, 20qr and 30rp = 2 × 5 = 10

Question (viii)
3x²y³, 10x³y², 6x²y²z
Solution:
3x²y³ = 3 × x × x × y × y × y
10x³y² = 2 × 5 × x × x × x × y × y
6x²y²z = 2 × 3 × x × x × y × y × z
∴ Common factors of 3x²y³, 10x³y² and 6x²y²z
= x × x × y × y = x²y²

2. Factorise the following expressions:

Question (i)
7x – 42
Solution:
7x = 7 × x
42 = 2 × 3 × 7
∴ 7 is common in both terms.
7x – 42 = 7 (x – 6)

Question (ii)
6p – 12q
Solution:
6p = 2 × 3 × p
12q = 2 × 2 × 3 × q
∴ 2 × 3 = 6 is common in both terms.
6p – 12q = 6 (p – 2q)

Question (iii)
7a² + 14a
Solution:
7a² = 7 × a × a
14a = 2 × 7 × a
∴ 7 × a = 7a is common in both terms.
∴ 7a² + 14a = 7a (a + 2)

Question (iv)
– 16z + 20z³
Solution:
– 16z = -2 × 2 × 2 × 2 × z
20z³ = 2 × 2 × 5 × z × z × z
∴ 2 × 2 × z = 4z is common in both terms.
∴ – 16z + 20z³ = 4z (- 4 + 5z²)

Question (v)
20l²m + 30alm
Solution:
= 2 × 2 × 5 × l × l × m + 2 × 3 × 5 × a × l × m
= 10lm (2l + 3a)

Question (vi)
5x²y – 15xy²
Solution:
= 5xy (x – 3y)

Question (vii)
10a² – 15b² + 20c²
Solution:
= 5 (2a² – 3b² + 4c²)

Question (viii)
– 4a² + 4ab – 4ca
Solution:
= 4a (- a + b – c)

Question (ix)
x²yz + xy²z + xyz²
Solution:
= xyz (x + y + z)

Question (x)
ax²y + bxy² + cxyz
Solution:
= xy (ax + by + cz)

3. Factorise:

Question (i)
x² + xy + 8x + 8y
Solution:
= x (x + y) + 8 (x + y)
= (x + y) (x + 8)

Question (ii)
15xy – 6x + 5y – 2
Solution:
= 3x (5y – 2) + 1 (5y – 2)
= (5y – 2) (3x + 1)

Question (iii)
ax + bx- ay – by
Solution:
= x (a + b) – y (a + b)
= (a+ b) (x- y)

Question (iv)
15pq + 15 + 9q + 25p
Solution:
= 15pq + 9q + 25p + 15
= 3q (5p + 3) + 5 (5p + 3)
= (5p + 3) (3q + 5)

Question (v)
z – 7 + 7xy – xyz
Solution:
= z – 7 – xyz + 7xy
= 1 (z – 7) – xy (z – 7)
= (z – 7) (1 – xy)