Find the nature of the roots of the following quartic function:

Question

$$x^4 + 2x^3 -13x^2 -14x + 24 = 0$$

I tried factoring, but it didn't seem to work out. I tried pulling out an $x^3$, but no progress. Please help.

Answer 1

You can start looking for the rational roots by the rational root theorem. Because the polynomial is monic, they'll need to be integers. We start with $\pm1$: $1$ is a root, but $-1$ isn't. We eventually find the roots are $1,\,-2,\,3,\,-4$.

Answer 2

$x=1$ is a root so divide by $x-1$ and try to factor the resulting cubic. $x=3$ is also a root so divide by $x-3$ and solve the resulting quadratic.

Answer 3

$$x^4 + 2x^3 -13x^2 -14x + 24=$$ $$=x^4-x^3+3x^3-3x^2-10x^2+10x-24x+24=(x-1)(x^3+3x^2-10x-24)=$$ $$=(x-1)(x^3+2x^2+x^2+2x-12x-24)=(x-1)(x+2)(x^2+x-12).$$ Can you end it now?