WBBSE Solutions For Class 9 Maths Chapter 8 Factorisation

Chapter 8 Factorisation Exercise 8.1

Let us factorise the following polynomials

Question 1. x3–3x+2

Solution: x3–3x+2

=x3−1−3x+3=(x)3−(1)3−3(x−1)=(x−1)(x2+x+1)−3(x−1)=(x−1)(x2+x+1−3)=(x−1)(x2+x−2)=(x−1)(x2+2x−x−2)=(x−1){x(x+2)−1(x+2)}=(x−1)(x+2)(x−1)=(x−1)2(x+2)

Question 2. x3 + 2x + 3

Solution: x3 + 2x + 3

=x3+1+2x+2=(x)3+(1)3+2(x+1)=(x+1)(x2−x+1)+2(x+1)=(x+1)(x2−x+1+2)=(x+1)(x2−x+3)

Question 3.a3–12a-16

Solution: a3–12a-16

=a3+8−12a−24=(a)3+(2)3−12(a+2)=(a+2)(a2−2a+4)−12(a+2)=(a+2)(a2−2a+4−12)=(a+2)(a2−2a−8)=(a+2)(a2−4a+2a−8)=(a+2){a(a−4)+2(a−4)}=(a+2)(a+2)(a−4)=(a+2)2(a−4)

Question 4. x3– 6x + 4

Solution: x3– 6x + 4

=x3−8−6x+12=(x)3−(2)3−6(x−2)=(x−2)(x2+2x+4)−6(x−2)=(x−2)(x2+2x+4−6)=(x−2)(x2+2x−2)

Question 5. x3–19x-30

Solution: x3–19x-30

=x3+8−19x−38=(x)3+(2)3−19(x+2)=(x+2)(x2−2x+4)−19(x+2)=(x+2)(x2−2x+4−19)=(x+2)(x2−2x−15)=(x+2)(x2−5x+3x−15)=(x+2){x(x−5)+3(x−5)}=(x+2)(x+3)(x−5)

Question 6. 4a3–9a2+ 3a +2

Solution: 4a3–9a2+ 3a +2
If a =1, 4a3–9a2+ 3a +2=0
∴ (a-1) is a factor of 4a3–9a2+ 3a +2

=4a3−9a2+3a+2=4a3−4a2−5a2+5a−2a+2=4a2(a−1)−5a(a−1)−2(a−1)=(a−1)(4a2−5a−2)