Question
Mathematics Question on What is Factorisation?
Factorise
- 4p2−9q2
- 63a2−112b2
- 49x2−36
- 16x5−144x3
- (l+m)2−(l−m)2
- 9x2y2−16
- (x2−2xy+y2)−z2
- 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)
= 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)][Usingidentitya2−b2=(a−b)(a+b)]
= (l+m−l+m)(l+m+l−m)
= 2mx2l=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][Usingidentity(a−b)2=a2−2ab+b2]
= [5a+(2b−7c)][5a−(2b−7c)][Usingidentitya2−b2=(a−b)(a+b)]
= (5a+2b−7c)(5a−2b+7c)