Find the integral roots of the polynomial $f(x) = x^3 + 6x^2 + 11x + 6$.

Given :

The given polynomial is $f(x) = x^3 + 6x^2 + 11x + 6$..

To find :

We have to find the integral roots of the polynomial $f(x) = x^3 + 6x^2 + 11x + 6$.

Solution :

$f(x) = x^3 +6x^2 +11x +6$

Here, $f(x)$ is a polynomial with integer coefficient and the coefficient of the highest degree term is 1. 

Therefore, integer roots of $f(x)$ are limited to the integer factors of $6$, which are  $\pm 1, \pm 2, \pm 3, \pm 6$

If $x = 1$,

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

$= 1+ 6+11+ 6$

$= 24$

$f(1) ≠ 0$

Therefore, $x = 1$ is not a zero of $f(x)$.

Similarly,

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

$=-1+6-11+6$

$=0$

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

$=-8+24-22+6$

$=0$

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

$=-27+54-33+6$

$=0$

Hence, the integral roots of \( f(x) \) are \( -1,-2,-3 \).

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Updated on: 10-Oct-2022

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