PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.2

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 14 Factorization Ex 14.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2

1. Factorise the following expressions:

Question (i)
a² + 8a + 16
Solution:
= (a)² + 2 (a)(4) + (4)²
= (a + 4)²

Question (ii)
p² – 10p + 25
Solution:
= (p)² – 2 (p)(5) + (5)²
= (P – 5)²

Question (iii)
25m² + 30m + 9
Solution:
= (5m)² + 2 (5m) (3) + (3)²
= (5m + 3)²

Question (iv)
49y² + 84yz + 36z²
Solution:
= (7y)² + 2 (7y)(6z) + (6z)²
= (7y + 6z)²

Question (v)
4x² – 8x + 4
Solution:
= 4(x² – 2x + 1)
= 4 [(x)² – 2 (x)(1) + (1)²]
= 4 (x – 1)²

Question (vi)
121b² – 88bc + 16c²
Solution:
= (11b)² – 2 (11b)(4c) + (4c)²
= (11b – 4c)²

Question (vii)
(l + m)² – 4lm [Hint: Expand (1 + m)² first]
Solution:
= l² + 2lm + m² – 4lm
= l² + 2lm – 4lm + m²
= l² – 2lm + m²
= (l)² – 2 (l) (m) + (m)²
= (l – m)²

Question (viii)
a⁴ + 2a²b² + b⁴
Solution:
= (a²)² + 2 (a²)(b²) + (b²)²
= (a² + b²)²

2. Factorise:

Question (i)
4p² – 9q²
Solution:
= (2p)² – (3q)²
= (2p – 3q) (2p + 3q)

Question (ii)
63a² – 112b²
Solution:
= 7 (9a² – 16b²)
= 7 [(3a)² -(4b)²]
= 7 (3a – 4b) (3a + 4b)

Question (iii)
49x² – 36
Solution:
= (7x)² – (6)²
= (7x – 6) (7x + 6)

Question (iv)
16x⁵ – 144x³
Solution:
= 16x³(x² – 9)
= 16x³ [(x)² – (3)²]
= 16x³ (x-3) (x + 3)

Question (v)
(l + m)² – (l – m)²
Solution:
=[(l + m) + (l – m)] [(l + m) – (l – m)]
= [l + m + l – m] [l + m – l + m]
= (2l) (2m)
= 4lm

Question (vi)
9x²y² – 16
Solution:
= (3xy)² – (4)²
= (3xy – 4) (3xy + 4)

Question (vii)
(x² – 2xy + y²) – z²
Solution:
= (x – y)² – (z)²
= [(x – y) – z] [(x – y) + z]
= (x – y – z) (x – y + z)

Question (viii)
25a² – 4b² + 28bc – 49c²
Solution:
= (25a²) – (4b² – 28bc + 49c²)
= (5a)² – (2b – 7c)²
= [(5a) – (2b – 7c)] [(5a) + (2b – 7c)]
= (5a – 2b + 7c) (5a + 2b – 7c)

3. Factorise the expressions:

Question (i)
ax² + bx
Solution:
= x (ax + b)

Question (ii)
7p² + 21q²
Solution:
= 7 (p² + 3q²)

Question (iii)
2x³ + 2xy² + 2xz²
Solution:
= 2x(x² + y² + z²)

Question (iv)
am² + bm² + bn² + an²
Solution:
= am² + bm² + an² + bn²
= m² (a + b) + n²(a + b)
= (a + b) (m² + n²)

Question (v)
(lm + l) + m + 1
Solution:
= l (m + 1) + 1 (m + 1)
= (m + 1) (l + 1)

Question (vi)
y(y + z) + 9(y + z)
Solution:
= (y + z)(y + 9)

Question (vii)
5y² – 20y – 8z + 2yz
Solution:
= 5y² – 20y + 2yz – 8z
= 5y (y – 4) + 2z (y – 4)
= (y- 4) (5y + 2z)

Question (viii)
10ab + 4a + 5b + 2
Solution:
= 2a (5b + 2) + 1 (5b + 2)
= (5b + 2) (2a + 1)

Question (ix)
6xy – 4y + 6 – 9x
Solution:
= 6xy – 4y – 9x + 6
= 2y (3x-2)-3(3x-2)
= (3x-2) (2y – 3)

4. Factorise:

Question (i)
a⁴ – b⁴
Solution:
= (a²)² – (b²)²
= (a² – b²) (a² + b²)
= ((a)² – (b²)] (a² + b²)
= (a – b) (a + b) (a² + b²)

Question (ii)
p⁴ – 81
Solution:
= (p²)² – (9)²
= (p² – 9) (p² + 9)
= ((p)² – (3)²] (p² + 9)
= (p – 3)(p + 3)(p² + 9)

Question (iii)
x⁴ – (y + z)⁴
Solution:
= (x²)² – (a²)² (∵ y + z = a)
= (x² – a²) (x² + a²)
= (x – a) (x + a) (x² + a²)
= [x – (y + z)] [x + (y + z)] [x² + (y + z)²] (∵ a = y + z)
= (x – y – z) (x + y + z) [x² + (y + z)²]

Question (iv)
x⁴ – (x – z)⁴
Solution:
= (x²)² – [(x – z)²]²
= [x² – (x – z)²] [x² + (x – z)²]
= [x² – (x² – 2xz + z²)] [x² + (x² – 2xz + z²)]
= (x² – x² + 2xz – z²) (x² + x² – 2xz + z²)
= (2xz – z²) (2x² – 2xz + z²)
= z (2x – z) (2x² – 2xz + z²)

Question (v)
a⁴ – 2a²b² + b⁴
Solution:
= (a²)² – 2(a²)(b²) + (b²)²
= (a² – b²)²
= (a² – b²) (a² – b²)
= (a – b) (a + b) (a – b) (a + b)

5. Factorise the following expressions:

Question (i)
p² + 6p + 8
Solution:
= p² + 6p + 9 – 1
= (p² + 6p + 9) – (1)
= (p + 3)² – (1)²
= (p + 3 + 1) (p + 3 – 1)
= (P + 4) (p + 2)
Here, last term is 8.
∴ 9 – 1 = 8.

OR
p² + 6p + 8
Here, ab = 8 and a + b = 6
On solving equations, a = 4, b = 2
Now, p² + 6p + 8
= p² + 4p + 2p + 8
= p (p + 4) + 2 (p + 4)
= (p + 4) (p + 2)

Question (ii)
q² – 10q + 21
Solution:
= q² – 10q + 25 – 4
= (q² – 10q + 25) – (4)
= (q – 5)² – (2)²
= (q – 5 + 2) (q – 5 – 2)
= (q – 3) (q – 7)
Here, last term is 21.
∴ 25 – 4 = 21.

OR
q² – 10q + 21
Here, ab = 21 and a + b = (- 10)
Possible values of a = 7 or (-7)
b = 3 or (- 3)
Let us check, 7 + 3 = 10 ≠ (- 10)
∴ a = – 7, b = – 3
Now, q² – 10q + 21
= q² – 7q – 3q + 21
= q (q – 7) – 3 (q – 7)
= (q – 7) (q – 3)

Question (iii)
p² + 6p – 16
Solution:
= p² + 6p + 9 – 25
= (P² + 6p + 9) – (25)
= (p + 3)² – (5)²
= (p + 3 – 5) (p + 3 + 5)
= (p – 2) (p + 8)
Here, last term is (-16).
∴ (-25) + 9 = (-16)

OR

p² + 6p – 16
Here, ab = – 16 and a + b = 6
Possible values of a = 8 or (-8) b = 2 or (-2)
Let us check, 8 + 2 = 10 ≠ 6
(- 8) + 2 = (-6) ≠ 6
8 + (-2) = 8 – 2 = 6
Now, p² + 6p – 16
= p² + 8p – 2p – 16
= P (P + 8) – 2 (p + 8)
= (p + 8) (p – 2)