Find the integral roots of the polynomial

Solution:

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

Clearly we can say that, the polynomial f(x) with an integer coefficient and the highest degree term coefficient which is known as leading factor is 1.

So, the roots of f(x) are limited to integer factor of 6, they are ±1, ± 2, ± 3, ± 6

Let x = -1

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

= -1 + 6 -11 + 6

= 0

Let x = – 2

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

= – 8 – (6 * 4) - 22 + 6

= – 8 + 24 - 22 + 6

= 0

Let x = – 3

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

= – 27 – (6 * 9) - 33 + 6

= – 27 + 54 - 33 + 6

= 0

But from all the given factors only -1, -2, -3 gives the result as zero.

So, the integral multiples of $x^{3}+6 x^{2}+11 x+6$ are $-1,-2,-3$.