KSEEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2

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KSEEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2

Question 1.
Factorize 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
= a2 + 4a + 4a + 16
= a(a + 4) + 4(a + 4)
= (a + 4) (a + 4)

(ii) p2 – 10p + 25
= p2 – 5p – 5p + 25
= p(p – 5) – 5(p – 5)
= (p – 5) (p – 5)

KSEEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2

(iii) 25m2 + 30m + 9
= (5m)2 + 2 (5m)(3) + (3)2
Since a2 + 2ab + b2 = (a + b)2
= (5m + 3)2
= (5m + 3) (5m + 3)

(iv) 4ay2 + 84yz + 36z2
= (7y)2 + 2(7y)(6z) + (6z)2
Since a2 + 2ab + b2 = (a + b)2
= (7y + 6z)2
= (7y + 6z) (7y + 6z)

(v) 4x2 – 8x + 4
= (2x)2 – 2(2x) (2) + (2)2
Since a2 – 2ab + b2 = (a – b)2
= (2x – 2)2
= (2x – 2) (2x – 2)

(vi) 121b2 – 88bc + 16c2
= (11b)2 – 2(11b) (4c) + (4c)2
Since a2 – 2ab + b2 = (a – b)2
= (11b – 4c)2
= (11b – 4c) (11b – 4c)

KSEEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2

(vii) (l + m)2 – 4lm
= l2 + 2lm + m2 – 4lm
= l2 – 2lm + m2
But a2 – 2ab + b2 = (a – b)2
(l – m)2 = (l – m) (l – m)

(viii) a4 + 2a2b2 + b4
= (a2)2 + 2a2b2 + (b2)2
But a2 + 2ab + b2 = (a + b)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) 4p2 – 9q2
= (2p)2 – (3q)2
= (2p – 3q) (2p + 3q)

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

KSEEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2

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

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

(v) (l + m)2 – (l – m)2
Let (l + m) = a, (l – m) = b
∴ (a2 – b2) = (a – b) (a + b)
= [(l + m) – (l – m)] [l + m + l – m]
= [l + m – l + m] [2l]
= [2m] [2l]
= 4lm

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

KSEEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2

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

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

Question 3.
Factorize 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 + b)

(ii) 7p2 + 21q2 = 7(p2 + 3q2)

KSEEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2

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

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

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

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

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

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

KSEEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2

(ix) 6xy – 4y + 6 – 9x
= 6xy – 9x – 4y + 6
= 3x (2y – 3) – 2(2y – 3)
= (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) a4 – b4
= (a2)2 – (b2)2
∴ (x2 – y2) = (x – y) (x + y)
= (a2 – b2) (a2 + b2)
= (a – b) (a + b) (a2 + b2)

(ii) p4 – 81
= (p2)2 – (9)2
∴ (a2 – b2) = (a – b) (a + b)
= (p2 – 9) (p2 + 9)
= (p2 – 32) (p2 + 9)
= (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 + y2 + z2 + 2yz]
= (x – y – z) (x + y + z) (x2 + y2 + z2 + 2yz)

KSEEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2

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

(v) a4 – 2a2b2 + b4
= (a2)2 – 2a2b2 + (b2)2
∴ (a2 – 2ab + b2) = (a – b)2
= (a2 – b2)2
= [(a – b) (a + b)]2
= (a – b)2 (a + b)2
= (a – b) (a – b) (a + b) (a + b)

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

KSEEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2

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

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