Using factor theorem, factorize of the polynomials:

Solution:

Given polynomial, $f(x)=x^{3}+6 x^{2}+11 x+6$

The constant term in f(x) is 6

The factors of 6 are ± 1, ± 2, ± 3, ± 6

Let, x + 1 = 0

=> x = -1

Substitute the value of x in f(x)

$f(-1)=(-1)^{3}+6(-1)^{2}+11(-1)+6$

= - 1 + 6 - 11 + 6

= 12 – 12

= 0

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

Similarly, (x + 2) and (x + 3) are also the factors 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 + 3)

$\Rightarrow x^{3}+6 x^{2}+11 x+6=k(x+1)(x+2)(x+3)$

$=x^{3}+6 x^{2}+11 x+6=k(x+1)(x+2)(x+3)$

Substitute x = 0 on both the sides

=> 0 + 0 + 0 + 6 = k(0 +1)(0 + 2)(0 + 3)

=>    6 = k(1*2*3)

=> 6 = 6k

=> k = 1

Substitute k value in f(x) = k(x + 1)(x + 2)(x + 3)

=> f(x) = (1)(x + 1)(x + 2)(x + 3)

=> f(x) = (x + 1)(x + 2)(x + 3)

$\therefore x^{3}+6 x^{2}+11 x+6=(x+1)(x+2)(x+3)$