Using factor theorem, factorize of the polynomials:

Solution:

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

The constant term in f(x) is -2

The factors of (-2) are ±1, ± 2

Let, x – 1 = 0

=> x = 1

Substitute the value of x in f(x)

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

= 1 + 2 – 1 – 2

= 0

Similarly, the other factors (x + 1) and (x + 2) of f(x)

Since, f(x) is a polynomial having a degree 3, it cannot have more than three linear factors.

∴ f(x) = k(x – 1)(x + 2)(x + 1 )

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

Substitute x = 0 on both the sides

0 + 0 – 0 – 2 = k(-1)(1)(2)

=> – 2 = - 2k

=> k = 1

Substitute k value in f(x) = k(x – 1)(x + 2)(x + 1)

f(x) = (1)(x – 1)(x + 2)(x + 1)

=> f(x) =  (x – 1)(x + 2)(x + 1)

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