Factorisation
NCERT Exercise 14.2
Q1: 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
Answer:
(i) a2 + 8a + 16 = (a)2 + 2 ☓ a ☓ 4 + (4)2
= (a + 4)2 [(x + y)2 = x2 + 2xy + y2]
(ii) p2 − 10p + 25 = (p)2 − 2 ☓ p ☓ 5 + (5)2
= (p − 5)2 [(a − b)2 = a2 − 2ab + b2]
(iii) 25m2 + 30m + 9 = (5m)2 + 2 ☓ 5m ☓ 3 + (3)2
= (5m + 3)2 [(a + b)2 = a2 + 2ab + b2]
(iv) 49y2 + 84yz + 36z2 = (7y)2 + 2 ☓ (7y) ☓ (6z) + (6z)2
= (7y + 6z)2 [(a + b)2 = a2 + 2ab + b2]
(v) 4x2 − 8x + 4 = (2x)2 − 2 (2x) (2) + (2)2
= (2x − 2)2 [(a − b)2 = a2 − 2ab + b2]
= [(2) (x − 1)]2 = 4(x − 1)2
(vi) 121b2 − 88bc + 16c2 = (11b)2 − 2 (11b) (4c) + (4c)2
= (11b − 4c)2 [(a − b)2 = a2 − 2ab + b2]
(vii) (l + m)2 − 4lm = l2 + 2lm + m2 − 4lm
= l2 − 2lm + m2
= (l − m)2 [(a − b)2 = a2 − 2ab + b2]
(viii) a4 + 2a2b2 + b4 = (a2)2 + 2 (a2) (b2) + (b2)2
= (a2 + b2)2 [(a + b)2 = a2 + 2ab + b2]
Q 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
Answer:
(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)
= 16 x2 [(x)2 − (3)2]
= 16 x2(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)]
[Using identity a2 − b2 = (a − b) (a + b)]
= (l + m − l + m) (l + m + l − m)
= 2m ☓ 2l
= 4ml
= 4lm
(vi) 9x2y2 − 16 = (3xy)2 − (4)2
=(3xy − 4) (3xy + 4) [a2 − b2 = (a − b) (a + b)]
(vii) (x2 − 2xy + y2) − z2 = (x − y)2 − (z)2 [(a − b)2 = a2 − 2ab + b2]
= (x − y − z) (x − y + z) [a2 − b2 = (a − b) (a + b)]
(viii) 25a2 − 4b2 + 28bc − 49c2 = 25a2 − (4b2 − 28bc + 49c2)
= (5a)2 − [(2b)2 − 2 ☓ 2b ☓ 7c + (7c)2]
= (5a)2 − [(2b − 7c)2]
[Using identity (a − b)2 = a2 − 2ab + b2]
= [5a + (2b − 7c)] [5a − (2b − 7c)]
[Using identity a2 − b2 = (a − b) (a + b)]
= (5a + 2b − 7c) (5a − 2b + 7c)
Q3: 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
Answer:
(i) ax2 + bx = a ☓ x ☓ x + b ☓ x = x(ax + b)
(ii) 7p2 + 21q2 = 7 ☓ p ☓ p + 3 ☓ 7 ☓ q ☓ q = 7(p2 + 3q2)
(iii) 2x3 + 2xy2 + 2xz2 = 2x(x2 + y2 + z2)
(iv) am2 + bm2 + bn2 + an2 = am2 + bm2 + an2 + bn2
= m2(a + b) + n2(a + b)
= (a + b) (m2 + n2)
(v) (lm + l) + m + 1 = lm + m + l + 1
= m(l + 1) + 1(l + 1)
= (l + l) (m + 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 = 10ab + 5b + 4a + 2
= 5b(2a + 1) + 2(2a + 1)
= (2a + 1) (5b + 2)
(ix) 6xy − 4y + 6 − 9x = 6xy − 9x − 4y + 6
= 3x(2y − 3) − 2(2y − 3)
= (2y − 3) (3x − 2)
Q4: Factorise
(i) a4 − b4
(ii) p4 − 81
(iii) x4 − (y + z)4
(iv) x4 − (x − z)4
(v) a4 − 2a2b2 + b4
Answer:
(i) a4 − b4 = (a2)2 − (b2)2
= (a2 − b2) (a2 + b2)
= (a − b) (a + b) (a2 + b2)
(ii) p4 − 81 = (p2)2 − (9)2
= (p2 − 9) (p2 + 9)
= [(p)2 − (3)2] (p2 + 9)
= (p − 3) (p + 3) (p2 + 9)
(iii) x4 − (y + 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]
= (x − y − z) (x + y + z) [x2 + (y + z)2]
(iv) 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 + (x − z)2]
= z(2x − z) [x2 + x2 − 2xz + z2]
= z(2x − z) (2x2 − 2xz + z2)
(v) a4 − 2a2b2 + b4 = (a2)2 − 2 (a2) (b2) + (b2)2
= (a2 − b2)2
= [(a − b) (a + b)]2
= (a − b)2 (a + b)2
Q5: Factorise the following expressions
(i) p2 + 6p + 8
(ii) q2 − 10q + 21
(iii) p2 + 6p − 16
Answer:
(i) p2 + 6p + 8
As we knkw, 8 = 4 ☓ 2 and 4 + 2 = 6
∴ p2 + 6p + 8 = p2 + 2p + 4p + 8
= p(p + 2) + 4(p + 2)
= (p + 2) (p + 4)
(ii) q2 − 10q + 21
It is seen that, 21 = (−7) ☓ (−3) and (−7) + (−3) = − 10
∴ q2 − 10q + 21 = q2 − 7q − 3q + 21
= q(q − 7) − 3(q − 7)
= (q − 7) (q − 3)
(iii) p2 + 6p − 16
It can be observed that, 16 = (−2) ☓ 8 and 8 + (−2) = 6
p2 + 6p − 16 = p2 + 8p − 2p − 16
= p(p + 8) − 2(p + 8)
= (p + 8) (p − 2)
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