Using factor theorem, factorize of the polynomials:
$x^{3}+6 x^{2}+11 x+6$
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)
$\therefore x^{3}+6 x^{2}+11 x+6=(x+1)(x+2)(x+3)$
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