Using factor theorem, factorize of the polynomials:

Solution:

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

The constant in f(x) is 20

The factors of 20 are ± 1, ± 2, ± 4, ± 5, ± 10, ± 20

Let, x + 1 = 0

=> x = -1

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

= -1 + 13 – 32 + 20

= 0

So, (x + 1) is the factor of f(x)

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

By, long division

$x^{2}+12 x+20$

$x+1, x^{3}+13 x^{2}+32 x+20$

$x^{3}+x^{2}$

(-)     (-)

$12 x^{2}+32 x$

 

$12 x^{2}+12 x$

(-)         (-)

20x – 20

20x – 20

(-)          (-)

0

$=>x^{3}+13 x^{2}+32 x+20=(x+1)\left(x^{2}+12 x+20\right)$

Now,

$x^{2}+12 x+20=x^{2}+10 x+2 x+20$

= x(x + 10) + 2(x + 10)

The factors are (x + 10) and (x + 2)

Hence, $x^{3}+13 x^{2}+32 x+20=(x+1)(x+10)(x+2)$