Maths Class 8 Factorization Chapter 12 Exercise 12.2 Question Answer

Solutions For All Chapters Maths Class 8

Ex 12.2 Class 8 Maths Question 1.
Factorise the following expressions.

(i) a² + 8a +16

(ii) p² – 10p + 25

(iii) 25m² + 30m + 9

(iv) 49y² + 84yz + 36z²

(v) 4x² – 8x + 4

(vi) 121b² – 88bc + 16c²

(vii) (l + m)² – 4lm. (Hint: Expand (l + m)² first)

(viii) a⁴ + 2a²b² + b⁴

Solution:

(i) a² + 8a + 16
Here, 4 + 4 = 8 and 4 × 4 = 16
a² + 8a +16
= a² + 4a + 4a + 4 × 4
= (a² + 4a) + (4a + 16)
= a(a + 4) + 4(a + 4)
= (a + 4) (a + 4)
= (a + 4)²

(ii) p² – 10p + 25
Here, 5 + 5 = 10 and 5 × 5 = 25
p² – 10p + 25
= p² – 5p – 5p + 5 × 5
= (p² – 5p) + (-5p + 25)
= p(p – 5) – 5(p – 5)
= (p – 5) (p – 5)
= (p – 5)²

(iii) 25m² + 30m + 9
Here, 15 + 15 = 30 and 15 × 15 = 25 × 9 = 225
25m² + 30m + 9
= 25m² + 15m + 15m + 9
= (25m² + 15m) + (15m + 9)
= 5m(5m + 3) + 3(5m + 3)
= (5m + 3) (5m + 3)
= (5m + 3)²

(iv) 49y² + 84yz + 36z²
Here, 42 + 42 = 84 and 42 × 42 = 49 × 36 = 1764
49y² + 84yz + 36z²
= 49y² + 42yz + 42yz + 36z²
= 7y(7y + 6z) +6z(7y + 6z)
= (7y + 6z) (7y + 6z)
= (7y + 6z)²

(v) 4x² – 8x + 4
= 4(x² – 2x + 1) [Taking 4 common]
= 4(x2 – x – x + 1)
= 4[x(x – 1) -1(x – 1)]
= 4(x – 1)(x – 1)
= 4(x – 1)²

(vi) 121b² – 88bc + 16c²
Here, 44 + 44 = 88 and 44 × 44 = 121 × 16 = 1936
121b² – 88bc + 16c²
= 121b² – 44bc – 44bc + 16c²
= 11b(11b – 4c) – 4c(11b – 4c)
= (11b – 4c) (11b – 4c)
= (11b – 4c)²

(vii) (l + m)² – 4lm
Expanding (l + m)², we get
l² + 2lm + m² – 4lm
= l² – 2lm + m²
= l² – Im – lm + m²
= l(l – m) – m(l – m)
= (l – m) (l – m)
= (l – m)²

(viii)  a⁴ + 2a²b² + b⁴

= a⁴ + a²b² + a²b² + b⁴

= a²(a² + b²) + b²(a² + b²)

= (a² + b²)(a² + b²)

= (a² + b²)²

Ex 12.2 Class 8 Maths Question 2.
Factorise.

(i) 4p² – 9q²

(ii) 63a² – 112b²

(iii) 49×2 – 36

(iv) 16x⁵ – 144x³

(v) (l + m)² – (l – m)²

(vi) 9x²y² – 16

(vii) (x² – 2xy + y²) – z²

(viii) 25a² – 4b² + 28bc – 49c²

Solution:

(i) 4p² – 9q²
= (2p)² – (3q)²
= (2p – 3q) (2p + 3q)
[∵ a2 – b2 = (a + b)(a – b)]

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

(iii) 49x² – 36 = (7x)² – (6)²
= (7x – 6) (7x + 6)
[∵ a² – b² = (a + b)(a – b)]

(iv) 16x⁵ – 144x³ = 16x³ (x2 – 9)
= 16x³ [(x)² – (3)²]
= 16x³(x – 3)(x + 3)
[∵ a2 – b² = (a + b)(a – b)]

(v) (l + m)² – (l – m)²
= (l + m) – (l – m)] [(l + m) + (l – m)]
[∵ a² – b² = (a + b)(a – b)]
= (l + m – l + m)(l + m + l – m)
= (2m) (2l)
= 4ml

(vi) 9x²y² – 16 = (3xy)v – (4)²
= (3xy – 4)(3xy + 4)
[∵ a² – b² = (a + b)(a – b)]

(vii) (x² – 2xy + y²) – z²
= (x – y)² – z²
= (x – y – z) (x – y + z)
[∵ a² – b² = (a + b)(a – b)]

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

Ex 12.2 Class 8 Maths Question 3.
Factorise the expressions.

(i) ax² + bx

(ii) 7p² + 21q²

(iii) 2x³ + 2xy² + 2xz²

(iv) am² + bm² + bn² + an²

(v) (lm + l) + m + 1

(vi) y(y + z) + 9(y + z)

(vii) 5y² – 20y – 8z + 2yz

(viii) 10ab + 4a + 5b + 2

(ix) 6xy – 4y + 6 – 9x

Solution:

(i) ax² + bx = x(ax + 5)

(ii) 7p² + 21q² = 7(p² + 3q²)

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

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

(v) (lm + l) + m + 1
= l(m + 1) + (m + 1)
= (m + 1) (l + 1)

(vi) y(y + z) + 9(y + z) = (y + z)(y + 9)

(vii) 5y² – 20y – 8z + 2yz
= 5y² – 20y + 2yz – 8z
= 5y(y – 4) + 2z(y – 4)
= (y – 4) (5y + 2z)

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

(ix) 6xy – 4y + 6 – 9x
= 6xy – 4y – 9x + 6
= 2y(3x – 2) – 3(3x – 2)
= (3x – 2) (2y – 3)

Ex 12.2 Class 8 Maths Question 4.
Factorise.

(i) a⁴  – b⁴

(ii) p⁴  – 81

(iii) x⁴  – (y + z)⁴

(iv) x⁴  – (x – z)⁴

(v) a⁴  – 2a²b² + b⁴

Solution:

(i) a⁴ – b⁴  – (a²)² – (b²)²

[∵ a² – b² = (a – b)(a + b)]
= (a² – b²) (a² + b²)
= (a – b) (a + b) (a² + b²)

(ii) p⁴  – 81 = (p²)² – (9)²
= (p² – 9) (p² + 9)
[∵ a² – b² = (a – b)(a + b)]
= (p – 3)(p + 3) (p² + 9)

(iii) x⁴ – (y + z)⁴ = (x²)² – [(y + z)²]²

[∵ a² – b² = (a – b)(a + b)]

= [x² – (y + z)²] [x² + (y + z)²]

= [x – (y + z)] [x + (y + z)] [x² + (y + z)²]

= (x – y – z) (x + y + z) [x² + (y + z)²]

(iv) x⁴ – (x – z)⁴ = (x²)² – [(y – z)²]²

= [x² – (y – z)²] [x² + (y – z)²]

= (x – y + z) (x + y – z) (x² + (y – z)²]

(v) a⁴ – 2a²b² + b⁴

= a⁴ – a²b² – a²b² + b⁴

= a²(a² – b²) – b²(a² – b²)

= (a² – b²)(a² – b²)

= (a² – b²)²

= [(a – b) (a + b)]²

= (a – b)² (a + b)²

Ex 12.2 Class 8 Maths Question 5.
Factorise the following expressions.

(i) p² + 6p + 8

(ii) q² – 10q + 21

(iii) p2 + 6p – 16

Solution:

(i) p² + 6p + 8
Here, 2 + 4 = 6 and 2 × 4 = 8
p² + 6p + 8
= p² + 2p + 4p + 8
= p (p + 2) + 4(p + 2)
= (p + 2) (p + 4)

(ii) q² – 10q + 21
Here, 3 + 7 = 10 and 3 × 7 = 21
q² – 10q + 21
= q² – 3q – 7q + 21
= q(q – 3) – 7(q – 3)
= (q – 3) (q – 7)

(iii) p² + 6p – 16
Here, 8 – 2 = 6 and 8 × 2 = 16
p² + 6p – 16
= p2 + 8p – 2p – 16
= p(p + 8) – 2(p + 8)
= (p + 8) (p – 2)