Question

Find the zeroes of the polynomials.
$x^3+6x^2+11x+6$

Answer 1

Let the given polynomial be
$p\left(x\right)=x^3+6x^2+11x+6$
The coefficient of the leading term is $1$ and the constant term is $6.$
Also the factors of $6$ are $1,2$ and $3.$ So the possible integral zeroes of $p\left(x\right)$ are $\pm 1,\pm 2$ and $\pm 3.$
Now, $p\left(x\right)$ does not have a negative coefficient of any term. So $p\left(x\right)$ cannot be zero for positive integral value of $x.$
Again
$p(-1)=(-1{)}^3+6(-1{)}^2+11(-1)+6$
$=-1+6-11+6=-12+12=0$
$\therefore -1$ is an integral zero of $p\left(x\right)$
$p(-2)=(-2{)}^3+6(-2{)}^2+11(-2)+6$
$=-8+24-22+6=30-30=0$
$\therefore -2$ is an integral zero of $p\left(x\right)$
Now
$p(-3)=(-3{)}^3+6(03{)}^2+11(-3)+6$
$=-27+54-33+6=60-60=0$
$\therefore -3$ is an integral zero of $p\left(x\right)$
Hence, the integral zeroes of the given polynomial are $-1,-2$ and $-3.$