Using factor theorem, factorize of the polynomials:

Solution:

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

The constant term is 2

The factors of 2 are ±1, ± 1/2

Let, x – 1= 0

=> x = 1

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

= 1 – 2 – 1 + 2

= 0

So, (x – 1) is the factor of f(x)

Divide f(x) with (x – 1) to get other factors

By, long division

$x^{2}-x-2$

$x-1 x^{3}-2 x^{2}-y+2$

$x^{3}-x^{2}$

(-)      (+)

$-x^{2}-x$

 

$-x^{2}+x$

(+)      (-)

– 2x + 2

– 2x + 2

(+)      (-)

0

$=x^{3}-2 x^{2}-y+2=(x-1)\left(x^{2}-x-2\right)$

Now,

$x^{2}-x-2=x^{2}-2 x+x-2$

= x(x – 2) + 1(x – 2)

=(x – 2)(x + 1)  are the factors

Hence, $x^{3}-2 x^{2}-y+2=(x-1)(x+1)(x-2)$