# How do you factor x^4 - x^3 - 5x^2 - x - 6?

Jun 27, 2016

Using the Rational Zeros Theorem the solution is $\left(x - 3\right) \left(x + 2\right) \left({x}^{2} + 1\right)$.

#### Explanation:

If you have a polynomial with integer coefficients you can try to find the solutions applying the Rational Zeros Theorem.
This theorem says that if a root is rational, it has to have the numerator as one of the factors of the constant and the denominator as one of the factors of the coefficient of the leading term.

It is easier to see in your case.
First of all your polynomial has the coefficients integers, they are

$\textcolor{red}{1} , - 1 , - 5 , - 1 , \textcolor{b l u e}{- 6}$.

The coefficient of the leading term is $\textcolor{red}{1}$, its possible factors are then $\textcolor{red}{\setminus \pm 1}$.
The constant term is $\textcolor{b l u e}{- 6}$ and the possible factors are $\textcolor{b l u e}{\setminus \pm 1 , \setminus \pm 2 , \setminus \pm 3 , \setminus \pm 6}$.

The theorem tells us that if a rational root exists it must be one of this

$\setminus \pm \frac{\textcolor{b l u e}{1}}{\textcolor{red}{1}} = \setminus \pm 1$
$\setminus \pm \frac{\textcolor{b l u e}{2}}{\textcolor{red}{1}} = \setminus \pm 2$
$\setminus \pm \frac{\textcolor{b l u e}{3}}{\textcolor{red}{1}} = \setminus \pm 3$
$\setminus \pm \frac{\textcolor{b l u e}{6}}{\textcolor{red}{1}} = \setminus \pm 6$

Then we try all of them and see if we obtain zero

${x}^{4} - {x}^{3} - 5 {x}^{2} - x - 6$
$x = 1$
$1 - 1 - 5 - 1 - 6 = - 12$ then $x = 1$ is not a root.
$x = - 1$
$1 + 1 - 5 + 1 - 6 = - 8$ then $x = - 1$ is not a root.
$x = 2$
${2}^{4} - {2}^{3} - 5 \cdot {2}^{2} - 2 - 6 = 20$ is not a root.
$x = - 2$
${2}^{4} + {2}^{3} - 5 \cdot {2}^{2} + 2 - 6 = 0$ IS A ROOT.
$x = 3$
${3}^{4} - {3}^{3} - 5 \cdot {3}^{2} - 3 - 6 = 0$ IS A ROOT.
$x = - 3$
${3}^{4} + {3}^{3} - 5 \cdot {3}^{2} + 3 - 6 = 60$ is not a root.
$x = 6$
${6}^{4} - {6}^{3} - 5 \cdot {6}^{2} - 6 - 6 = 888$ is not a root.
$x = - 6$
${6}^{4} + {6}^{3} - 5 \cdot {6}^{2} + 6 - 6 = 1332$ is not a root.

We were lucky and we found two roots, $- 2$ and $3$.
Then we can start our factorization as
${x}^{4} - {x}^{3} - 5 {x}^{2} - x - 6 = \left(x - 3\right) \left(x + 2\right) \left(a {x}^{2} + b x + c\right)$

The part that is missing is the quadratic polynomial $a {x}^{2} + b x + c$.
If we multiply the terms we have

$\left({x}^{2} - x - 6\right) \left(a {x}^{2} + b x + c\right)$

$a {x}^{4} + b {x}^{3} + c {x}^{2} - a {x}^{3} - b {x}^{2} - c x - 6 a {x}^{2} - 6 b x - 6 c$

$a {x}^{4} + \left(b - a\right) {x}^{3} + \left(c - 6 a - b\right) {x}^{2} - \left(c + 6 b\right) x - 6 c$

if we compare this with the initial polynomial

${x}^{4} - {x}^{3} - 5 {x}^{2} - x - 6$

we have

$a = 1 , b - a = - 1 , c - 6 a - b = - 5 , c + 6 b = 1 , 6 c = 6$

$a = 1 , b = 0 , c = 1$.

Then the final factorization is

$\left(x - 3\right) \left(x + 2\right) \left({x}^{2} + 1\right)$.

If you want to factorize in the complex numbers you can write

$\left(x - 3\right) \left(x + 2\right) \left(x + i\right) \left(x - i\right)$.