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

PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.1

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.]

PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.1

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

PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.1

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)

PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.1

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)

PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.1

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)

PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.1

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)

PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.1

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)

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