MP Board Class 8th Maths Solutions Chapter 14 Factorization Ex 14.2

Question 1.
Factorise the following expressions.
(i) a2 + 8o + 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) The expression is a2 + 8a +16
= a2 + 2 × 4 × a + (4)2
= (a + 4)2 = (a + 4) (a + 4)

(ii) The expression is p2 – 10p + 25
= (p)2 – 2 × 5 × p + (5)2
= (P – 5)2 = (p – 5)(p – 5)

(iii) The expression is 25m2 + 30m + 9
= (5m)2 + 2 × 3 × (5m) + (3)2 = (5m + 3)2
= (5m + 3) (5m + 3)

(iv) The expression is 49y2 + 84yz + 36z2
= (7y)2 + 2 × (7y) × (6z) + (6z)2 = (7y + 6z)2
= (7y + 6z)(7y + 6z)

(v) The expression is 4x2 – 8x + 4
= (2x)2 – 2 × (2x) × 2 + (2)2 = (2x – 2)2
= [2(x – 1)]2 = 4(x – 1)2
= 4(x – 1)(x – 1)

(vi) The expression is 121b2 – 88bc + 16c2
= (11b)2 – 2 × (11b) (4c) + (4c)2 = (11b – 4c)2
= (11b – 4c)(11b – 4c)

(vii) The expression is (l + m)2 – 4lm
= l2 + 2 × l × m + m2 – 4lm [ ∵ (a + b)2 = a2 + 2ab + b2]
= l2 + 2lm + m2 – 4lm = l2 – 2lm + m2
= (l – m)2 = (l – m)(l – m)

(viii)The expression is a4 + 2a2b2 + b4
= (a2)2 + 2 × a2 × 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.
Solution:
(i) The expression is 4p2 – 9q2
= (2P)2 – (3q)2 = (2p + 3q) (2p – 3q)

(ii) The expression is 63a2 – 112b2
= 7[9a2 – 16b2] = 7[(30)2 – (4b)2]
= 7(3o + 4b) (3a – 4b).

(iii) The expression is 49x2 – 36 = (7x)2 – (6)2
= (7x + 6) (7x – 6).

(iv) The expression is 16x5 – 144x3
= 16x3(x2 – 9) = 16x3 (x2 – 32)
= 16x3 (x + 3)(x – 3).

(v) The expression is (l + m)2 – (l – m)2
= (l2 + 2lm + m2) – (l2 – 2lm + m2)
= l2 + 2lm + m2 – l2 + 2lm – m2 = 4lm

(vi) The expression is 9x2y2 – 16
= (3xy)2 – (4)2 = (3xy + 4) (3xy – 4)

(vii) The expression is (x2 – 2xy + y2) – z2
= (x – y)2 – z2 = (x – y + z) (x – y – z)

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

MP Board Class 8th Maths Solutions Chapter 14 Factorization Ex 14.2

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) The expression is ax2 + bx = x(ax + b)
(ii) The expression is 7p2 + 21q2 = 7(p2 + 3q2 )
(iii) The expression is 2x3 + 2xy2 + 2xz2
= 2x(x2 + y2 + z2 ).

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

(v) The expression is (lm + l) + m + 1
= l(m + 1) + 1 (m + 1)= (l + 1) (m + 1)

(vi) The expression is y(y + z) + 9(y + z)
= (y + 9) (y + z).

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

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

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

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) The expression is a4 – b2
= (a2)2 – (b2)2
= (a2 + b2) (a2 – b2)
= (a2 + b2)(a + b)(a – b).

(ii) The expression is p4 – 81 = (p)4 – (3)4
= (P2)2 – (32)2 = (p2 + 32) (p2 – 32)
= (p2 + 9) (p + 3) (p – 3)

(iii) The expression is x4 – (y + z)4
= (x2)2 – ((y + z)2)2
= [x2 + (y + z)2] [x2 – (y + z)2]
= [x2 + (y + z)2] (x + y + z) (x – (y + z))
= [x2 + (y + z)2] (x + y + z) (x – y – z)

(iv) The expression is x4 – (x – z)4
= (x2)2 – ((x – z)2)2
= (x2 – (x – z)2)(x2 + (x – z)2)
= (x – x + z)(x + x – z)(x2 + x2 + z2 – 2xz)
= z(2x – z) (2x2 – 2xz + z2).

(v) The expression is a4 – 2a2b2 + b4
= (a2)2 – 2(a2) (b2) + (b2)2 = (a2 – b2)2
= [(a + b) (a – b)]2 = (a + b)2 (a – b)2

Question 5.
Factorise the following expressions.
(i) p2 + 6p + 8
(ii) q2 – 10q + 21
(iii) p2 + 6p – 16
Solution:
(i) The expression is p2 + 6p + 8
= p2 + 4p + 2p + 8 = p(p + 4) + 2 (p + 4)
= (p + 2) (p + 4)

(ii) The expression is q2 – 10q + 21
= q2 – 7q – 3q + 21 = q(q – 7) – 3 (q – 7)
= (q – 3) (q – 7)

(iii) The expression is p2 + 6p – 16
= p2 + 8p – 2p – 16
= p(p + 8) – 2(p + 8) = (p – 2) (p + 8)

MP Board Class 8th Maths Solutions

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