# Factorise:(i) $x^{3}-2 x^{2}-x+2$(ii) $x^{3}-3 x^{2}-9 x-5$(iii) $x^{3}+13 x^{2}+32 x+20$(iv) $2 y^{3}+y^{2}-2 y-1$

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To do:

We have to factorise the given expressions.

Solution:

(i) $x^{3}-2 x^{2}-x+2$

Let $f(x)=x^{3}-2 x^{2}-x+2$

Factors of $2$ are $\pm 1$ and $\pm 2$.     [Product of coefficient of $x^3$ and constant term $=1\times2=2$]

By trial and error method, we get,

$f(1)=(1)^{3}-2 (1)^{2}-(1)+2$

$=1-2(1)-1+2$

$=3-3$

$=0$

This implies,

$x-1$ is a factor of $f(x)$.

Therefore,

$f(x)=x^3-2x^2-x+2$

$= x^3 - x^2 - x^2 + x - 2x + 2$

$= x^2 (x - 1) - x (x - 1)-2 (x - 1)$

$= (x - 1)(x^2 - x - 2)$

$= (x - 1)(x^2 - 2x+x-2)$

$= (x - 1) [x (x - 2) + 1 (x - 2)]$

$= (x - 1) (x - 2)(x + 1)$

(ii) $x^{3}-3 x^{2}-9 x-5$

Let $f(x)=x^{3}-3 x^{2}-9 x-5$

Factors of $-5$ are $\pm 1$ and $\pm 5$.     [Product of coefficient of $x^3$ and constant term $=1\times-5=-5$]

By trial and error method, we get,

$f(5)=(5)^{3}-3 (5)^{2}-9(5)-5$

$=125-3(25)-45-5$

$=125-75-50$

$=125-125$

$=0$

This implies,

$x-5$ is a factor of $f(x)$.

Therefore,

$f(x)=x^3-3x^2-9x-5$

$= x^3-5x^2 + 2x^2-10x+x-5$

$= x^2(x - 5)+2x(x - 5)+1(x - 5)$

$= (x - 5) (x^2 + 2x + 1)$

$= (x - 5) (x^2 + x + x + 1)$

$= (x - 5) [x (x + 1)+ 1 (x+ 1)]$

$= (x - 5) (x + 1) (x + 1)$

$= (x - 5)(x+1)^2$

(iii) $x^{3}+13 x^{2}+32 x+20$

Let $f(x)=x^{3}+13 x^{2}+32 x+20$

Factors of $20$ are $\pm 1, \pm 2 \pm 4, \pm 5, \pm 10$ and $\pm 20$.     [Product of coefficient of $x^3$ and constant term $=1\times20=20$]

By trial and error method, we get,

$f(-1)=(-1)^{3}+13 (-1)^2+32(-1)+20$

$=-1+13(1)-32+20$

$=-33+13+20$

$=33-33$

$=0$

This implies,

$x-(-1)=x+1$ is a factor of $f(x)$.

Therefore,

$f(x)=x^{3}+13 x^{2}+32 x+20$

$= x^3+ x^2 + 12x^2 + 12x+ 20x+ 20$

$=x^2(x+ 1) + 12x(x+ 1)+ 20 (x+ 1)$

$= (x+1)(x^2+12x+20)$

$= (x+ 1) (x^2+ 10x + 2x+ 20)$

$= (x+1)[x(x+10)+2(x+10)]$

$= (x+ 1) (x+ 10) (x + 2)$

(iv) $2 y^{3}+y^{2}-2 y-1$

Let $f(y)=2 y^{3}+y^{2}-2 y-1$

Factors of $-2$ are $\pm 1$ and $\pm 2$.     [Product of coefficient of $y^3$ and constant term $=2\times-1=-2$]

By trial and error method, we get,

$f(1)=2(1)^{3}+ (1)^2-2(1)-1$

$=2+1-2-1$

$=3-3$

$=0$

This implies,

$y-1$ is a factor of $f(y)$.

Therefore,

$f(y)=2 y^{3}+y^{2}-2 y-1$

$= 2y^3 - 2y^2+ 3y^2 - 3y + y - 1$

$= 2y^2(y - 1) + 3y(y - 1)+1(y - 1)$

$= (y - 1) (2y^2 + 3y + 1)$

$= (y - 1)(2y^2 + 2y +y+1)$

$= (y - 1) [2y (y + 1) + 1 (y + 1)]$

$= (y - 1)(y+1)(2y+1)$

Updated on 10-Oct-2022 13:39:07