Using factor theorem, factorize of the polynomials:

Solution:

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

The constant in f(x) is -5

The factors of -5 are ±1, ±5

Let, x + 1 = 0

=> x = -1

$f(-1)=(-1)^{3}-3(-1)^{2}-9(-1)-5$

= -1 – 3 + 9 – 5

= 0

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

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

By, long division

$x^{2}-4 x-5$

$x+1 x^{3}-3 x^{2}-9 x-5$

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

(-)   (-)

$-4 x^{2}-9 x$

 

$-4 x^{2}-4 x$

(+)      (+)

- 5x – 5

- 5x – 5

(+)     (+)

0

$=x^{3}-3 x^{2}-9 x-5=(x+1)\left(x^{2}-4 x-5\right)$

Now,

$x^{2}-4 x-5=x^{2}-5 x+x-5$

= x(x – 5) + 1(x – 5)

The factors are (x – 5) and (x + 1)

Hence, $x^{3}-3 x^{2}-9 x-5=(x+1)(x-5)(x+1)$