# NCERT Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2

These NCERT Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 8 Maths Chapter 14 Factorization Exercise 14.2

Question 1.
Factorise the following expressions.
(i) a2 + 8a + 16
(ii) p2 – lOp + 25
(iii) 25m2 + 30m+ 9
(iv) 49y2 + 84yz + 36z2
(v) 4x2 – 8x + 4
(vi) 121b2 – 88bc + 16c2
(vii) (1 + m)2 – 4lm
(viii) a4 + 2a2b2 + b4
(Hint: Expand (1 + m)2 first)
(i) a2 + 8a + 16
= a2 + 2 x a x 4 + 42
= (a + 4)2
= (a + 4) (a + 4)
(ii) p2 – 10p + 25
= p2 – 2 x 5p + 52– (p – 5)2
= (p – 5) (p – 5)

(iii) 25m2 + 30m + 9
= (5m)2 + 2 x 5m x 3 + 32
= (5m + 3)2
= (5m + 3) (5m +3)

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

(v) 4x2 – 8x + 4
= (2x)2 – 2 x 2x x 2 + 22
= (2x – 2)2 = 22 (x – 1)2
= 4 (x – 1) (x – 1)

(vi) 12lb2 – 88bc + 16c2
= (11b)2 – 2 x 11b x 4c + (4c)2
= (11b – 4c)2
= (11b – 4c) (11b – 4c)

(vii) (l + m)2 – 4lm
= l2 + 2lm + m2 – 4lm
= l2 + (21m – 41m) + m2
= 1 – 21m + m2
= (1 – m)2 = (1 – m) (1 – m)

(viii) a4 + 2a2 b2 + b4
= (a2)2 + 2 x a2 x b2 + (b2)2
= (a2 + b2)2
= (a2 + b2) (a2 + b2)

Question 2.
Factorise
(i) 4p2 – 9q2
(ii) 63a2 – 112b2
(iii) 49x2 – 36
(iv) 16x5 – 144x3
(v) (l + m)2 – (l – m)2
(vi) 9x2y2 – 16
(vii) (x2 – 2xy + y2) – z2
(viii) 25a2 – 4b2 + 28bc – 49c2
(i) 4p2 – 9q2
= (2p)2 – (3q)2
= (2p + 3q) (2p – 3q)
[a2 – b2 = (a + b) (a – b)]

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

(iii) 49x2 – 36
= (7x)2 – 62 = (7x + 6) (7x – 6)
[Using a2 – b2 = (a + b) (a – b)]

(iv) 16x3 – 144x3
= x3 [16×2 – 144]
= x3 x 16 [x2 – 9]
= 16x3 [x2 – 32]
= 16x3 (x + 3) (x – 3)

(v) (l + m)2 – (l – m)2
= (l + m + l – m) [l + m – (l – m)]
= 2l [l + m – l + m]
= 2l x 2m = 4lm

(vi) 9x2y2 – 16
= (3xy)2 – 42
= (3xy + 4) (3xy – 4)

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

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

Question 3.
Factorise the expressions.
(i) ax2 + bx
(ii) 7p2 + 21q2
(iii) 2x3 + 2xy2 + 2xz2
(iv) am2 + bm2 + bn2 + an2
(v) (lm + 1) + m + 1
(vi) y (y + z) + 9 (y + z)
(vii) 5y2 – 20y – 8z + 2yz
(viii) 10ab + 4a + 5b + 2
(ix) 6xy – 4y + 6 – 9x
(i) ax2 + bx
= a x x x x + b x x
= x (ax + b)

(ii) 7p2 + 21q2
= 7 x p x p + 3 x 7 x q x q – 7 (p2 + 3q2)

(iii) 2x3 + 2xy2 + 2xz2
= 2 x x x x x x + 2 x x x y x y + 2 x x x z x z
= 2x(x xx + yxy + zxz)
= 2x (x2 + y2 + z2)

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

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

(vi) y (y + z) + 9 (y + z)
= (y + z) (y + 9)
[(y + z) is common for both terms)]

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

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

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

Question 4.
Factorise
(i) a4 – b4
(ii) p4 – 81
(iii) x4 – (y + z)4
(iv) x24 – (x – z)
(v) a4 – 2a2b2 + b4
(i) a4 – b4
= (a2)2 – (b2)2
[(using a2 – b2 = (a + b) (a – b)]
= (a2 + b2) (a2 – b2)
= (a2 + b2) (a + b) (a – b)

(ii) p4 – 81
= (p2)2 – 92
[Using a2 – b2 = (a + b) (a – b)]
= (p2 + 9) (p2 – 9)
= (P2 + 9) [p2 – 32 ]
= (p2 + 9) (p + 3) (p – 3)

(iii) x2 – (y + z)4
= (x2)2 – [(y + z)2]2
[using a2 – b2 = (a + b) (a – b)
= [x2 + (y + z)2] [x2 – (y + z)2]
= [x2 + (y + z2)] (x + y + z) [x – (y + z)
= [x2 + (y + z)2] [x + y + z) (x – y – z)

(iv) x4 – (x – z)4
= (x2)2 – [(x-z)2]2
= [x2 + (x – z)2] [x2 – (x – z)2]
[Using a2 – b2 = (a + b) (a – b)]
= [x2 + (x – z)2 ] [(x + x – z) (x – x + z)]
= [x2 + (x – z)2] (2x – z) (z)
= (x2 + x2 + z2 – 2xz) (2x – z) (z)
= (2x2 + z2 – 2xz) (2x – z) (z)
= z (2x – z) (2x2 + z2 – 2xz)

(v) a4 – 2a2 b2 + b4
= (a2)2 – 2 x a2 x b2 + (b2)2 = (a2 – b2 )2
= (a2 – b2) x (a2 – b2)
= (a + b) (a – b) (a + b) (a – b)
= (a + b)2 (a – b)2

Question 5.
Factorise the following expressions.
(i) p2 + 6p + 8
(ii) q2 – lOq + 21
(iii) p2 + 6p – 16
(i) p2 + 6p + 8
= p2 + 4p + 2p + 8
[8 = 2 x 4 and 2 + 4 = 6]
= p (p + 4) + 2 (p + 4)
= (p + 4) (p + 2)

(ii) q2 – 10q + 21
= q2 – 7q – 3q + 21
[21 = – 7 x – 3 and – 10 = – 7 – 3]
= q(q – 7) – 3(q – 7)
= (q – 7) (q – 3)

(iii) p2 + 6p – 16
= p2 + 8p – 2p – 16
[-16 = 8 x – 2 and 8 – 2 = 6]
= p(p + 8) – 2 (p + 8)
= (p + 8) (p – 2)2

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