Factorisation Using Algebraic Identities
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Factorise the following expression:
Factorise the following expression: .
Factorise:
a6−7a3−8
Factorise:
4x4−x2−12x−36
Factorise:
8a3−b3−4ax+2bx
Factorise:
(i) x3−2x2−x+2
(ii) x3−3x2−9x−5
(iii) x3+13x2+32x+20
(iv) 2y3+y2−2y−1
6(2a+3b)2−8(2a+3b)
Prove that 0.85×0.85×0.85+0.15×0.15×0.150.85×0.85−0.85×0.15+0.15×0.15=1.
Factorise:
a(a−1)−b(b−1)
Factorise:
a2−(2a+3b)2
Factorise:
9a2+3a−8b−64b2
The factors of x3−1+y3+3xy are
(x−1+y)(x2+1+y2+x+y−xy)
(x+y+1)(x2+y2+1−xy−x−y)
(x−1+y)(x2−1−y2+x+y+xy)
3(x+−1)(x2+y2−1)
Factorise:
4x2+14x2+1
Factorise:
a−b−a3+b3
Factorise:
a3−27b3+2a2b−6ab2
Show that:
(i) 133−53 is divisible by 8.
(ii) 353+273 is divisible by 62.
Prove that 59×59×59−9×9×959×59+59×9+9×9=50.
Factorise:
4a2−12a+9−49b2
Factorise:
(a2−1)(b2−1)+4ab
Factorised form of 23xy – 46x + 54y – 108 is
(a) (23x + 54) (y – 2)
(b) (23x + 54y) (y – 2)
(c) (23xy + 54y) (– 46x – 108)
(d) (23x + 54) (y + 2)
Question 32
Factorise the following:
(i) 1–64a3–12a+48a2
(ii) 8p3+125p2+625p+1125
Question 36
Factorise:
(i) a3−8b3−64c3−24abc
(ii) 2√2a3+8b3−27c3+18√2abc
Factorise:
4a2−49b2+2a−7b
Factorise: 4xy−x2−4y2+z2
Factorise:
4a2−(4b2+4bc+c2)
Factorise:
a2+b2−c2−d2+2ab−2cd
Factorise:
x4+y4−3x2y2
Factorise:
a3−27a3
Factorise:
a2−81(b−c)2
x3+13x2+32x+20
- (x+1)(x+2)(x+10)
- (x−1)(x+2)(x−10)
- (x−1)(x−2)(x+10)
- (x+1)(x−2)(x−10)