# How do you factor r^4 + r^3 − 3r^2 − 5r − 2?

Jan 28, 2016

${r}^{4} + {r}^{3} - 3 {r}^{2} - 5 r - 2 = {\left(r + 1\right)}^{3} \left(r - 2\right)$

#### Explanation:

One way to do this is finding one root of the term and then performing polynomial long division. The procedure can be repeated until there is only a quadratic term left.

1) Searching for the first root / factor

If searching for a root, it is generally a good idea to evaluate the term for values like $r = 1$, $r = - 1$, $r = 2$, $r = - 2$, ...

Here, it works with $r = - 1$:

${\left(- 1\right)}^{4} + {\left(- 1\right)}^{3} - 3 {\left(- 1\right)}^{2} - 5 \cdot \left(- 1\right) - 2 = 1 - 1 - 3 + 5 - 2 = 0$

Thus, $r = - 1$ is a root and $\left(r - \left(- 1\right)\right)$ is one of the factors of your term.

2) Polynomial long division

Let's use divide by $r + 1$ to make the term easier and then hopefully find further roots.

$\textcolor{w h i t e}{\times} \left({r}^{4} + {r}^{3} - 3 {r}^{2} - 5 r - 2\right) \div \left(r + 1\right) = {r}^{3} - 3 r - 2$
$- \left({r}^{4} + {r}^{3}\right)$
color(white)(x) color(white)(xxxxxx)/
$\textcolor{w h i t e}{\times \times \times} 0 - 3 {r}^{2} - 5 r$
$\textcolor{w h i t e}{\times \times} - \left(- 3 {r}^{2} - 3 r\right)$
color(white)(xxxxxx) color(white)(xxxxxxxxx)/
$\textcolor{w h i t e}{\times \times \times \times \times x} - 2 r - 2$
$\textcolor{w h i t e}{\times \times \times \times x} - \left(- 2 r - 2\right)$
color(white)(xxxxxxxxxxx) color(white)(xxxxxxxx)/
$\textcolor{w h i t e}{\times \times \times \times \times \times \times \times x} 0$

${r}^{4} + {r}^{3} - 3 {r}^{2} - 5 r - 2 = \left(r + 1\right) \left({r}^{3} - 3 r - 2\right)$

3) Searching for the second root / factor

Now, let's try and factor ${r}^{3} - 3 r - 2$ further.

We can repeat the same procedure as before and again, we find that $r = - 1$ is a root:

${\left(- 1\right)}^{3} - 3 \cdot \left(- 1\right) - 2 = - 1 + 3 - 2 = 0$

4) Polynomial long division

Thus, we can divide $\left({r}^{3} - 3 r - 2\right)$ by $\left(r + 1\right)$ to simplify the term further:

$\textcolor{w h i t e}{\times} \left({r}^{3} \textcolor{w h i t e}{\times \times} - 3 r - 2\right) \div \left(r + 1\right) = {r}^{2} - r - 2$
$- \left({r}^{3} + {r}^{2}\right)$
 color(white)(x) color(white)(xxxxxx) /
$\textcolor{w h i t e}{\times \times} - {r}^{2} - 3 r$
$\textcolor{w h i t e}{\times} - \left(- {r}^{2} - r\right)$
 color(white)(xxxx) color(white)(xxxxxxxx) /
$\textcolor{w h i t e}{\times \times \times \times} - 2 r - 2$
$\textcolor{w h i t e}{\times \times \times} - \left(- 2 r - 2\right)$
 color(white)(xxxxxxxx) color(white)(xxxxxxxx) /
$\textcolor{w h i t e}{\times \times \times \times \times \times \times} 0$

At this point, we can factor the term as follows:

${r}^{4} + {r}^{3} - 3 {r}^{2} - 5 r - 2 = \left(r + 1\right) \left(r + 1\right) \left({r}^{2} - r - 2\right)$

At this point, the only thing left to do is factoring the term ${r}^{2} - r - 2$.

There are a lot of ways to do that. Let me show you one of my favourites.

Basically, you would like to have something like this:

${r}^{2} - r - 2 = \left(r + a\right) \left(r + b\right)$

$= {r}^{2} + \left(a + b\right) r + a \cdot b$

Thus, you need to find $a$ and $b$ so that $a + b = - 1$ and $a \times b = - 2$

The solution to this is $a = 1$ and $b = - 2$.

${r}^{2} - r - 2 = \left(r + 1\right) \left(r - 2\right)$
${r}^{4} + {r}^{3} - 3 {r}^{2} - 5 r - 2 = \left(r + 1\right) \left(r + 1\right) \left(r + 1\right) \left(r - 2\right)$
$= {\left(r + 1\right)}^{3} \left(r - 2\right)$