Question

Factorise :
(i) $x^3$−2$x^2$−x+2
(ii) 2$y^3$+$y^2$−2y−1

Answer 1

(i) Let p(x)=$x^3$−2$x^2$−x+2

Factors of constant term 2 are 1 and 2.

Let's check if (x−1) is a factor of p(x).

p(1)=$1^3$−2${\left(1\right)}^2$−1+2 =1−2−1+2=0

By Factor theorem, we can say that (x−1) is the factor of p(x)=$x^3$−2$x^2$−x+2

We can also find other factors of p(x) by doing long division but we will just split the terms to take (x−1) common from p(x).

p(x)=$x^3$−2$x^2$−x+2

=$x^3$−$x^2$−$x^2$+x−2x+2

=$x^2$(x−1)−x(x−1)−2(x−1)

=(x−1)($x^2$−x−2)

=(x−1)($x^2$−2x+x−2)

=(x−1)[x(x−2)+1(x−2)]

=(x−1)(x−2)(x+1)

(ii) Let p(y)=2$y^3$+$y^2$−2y−1

Factor of constant term 1 is 1.

Let's check if (y−1) is a factor of p(y).

p(1) = 2${\left(1\right)}^3$+${\left(1\right)}^2$−2(1)−1 = 2+1−2−1 = 3−3 = 0

Therefore, by factor theorem, we can say that (y−1) is a factor of p(y).

Now, we will split the terms accordingly to take (y−1) common from p(y). We can also do long division to find other factors but it may take more time.

p(y) = 2$y^3$+$y^2$−2y−1

= 2$y^3$+2$y^2$−$y^2$−y−y−1

=2$y^2$(y+1)−y(y+1)−1(y+1)

=(y+1)(2$y^2$−y−1)

=(y+1)(2$y^2$−2y+y−1)

=(y+1)[2y(y−1)+1(y−1)]

=(y+1)(y−1)(2y+1)