Question

Find the integral roots of the polynomial $x^3+6x^2+11x+6$

Answer 1

Let $p\left(x\right)=x^3+6x^2+11x+6$

We shall now can look all the factors of $6$. Some of these are $\pm 1,\pm 2,\pm 3,\pm 6$

Put $x=–1$

$p(–1)=(–1{)}^3+6(–1{)}^2+11(–1)+6=–1+6–11+6=0$

By trial method, we find that $p(-1)=0.$ So, $(x+1)$ is a factor of $p\left(x\right).$

$p\left(x\right)=x^3+6x^2+11x+6$

$=x^3+x^2+5x^2+5x+6x+6$ [We can write, $6x^2$ as $5x^2+x^2$ and $11x$ as $5x+6x$]

$=x^2(x+1)+5x(x+1)+6(x+1)$

$=(x+1)(x^2+5x+6)$ [Taking $(x+1)$ common]

Now, we can factorise $(x^2+5x+6)$ by middle term method as

$(x^2+5x+6)$

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

$=x(x+2)+3(x+2)$

$=(x+2)(x+3)$

$\therefore p\left(x\right)=(x+1)(x^2+5x+6)=(x+1)(x+2)(x+3)$

To find the roots, we have to equate $p\left(x\right)=0$

$\therefore (x+1)(x+2)(x+3)=0$

$\Rightarrow (x+1)=0or,(x+2)=0or,(x+3)=0$

$\therefore x=-1,-2,-3$

Hence, the integral roots (means roots which are integers) of the given polynomial are $-1,-2,-3.$