# How do you factor  x^4 + 2x^3 - 8x -16?

Jun 27, 2018

$\left(x + 2\right) \left(x - 2\right) \left({x}^{2} + 2 x + 4\right)$

#### Explanation:

${x}^{4} + 2 {x}^{3} - 8 x - 16$

$\therefore = {x}^{3} \left(x + 2\right) - 8 \left(x + 2\right)$

$\therefore = \left(x + 2\right) \left({x}^{3} - 8\right)$

$\therefore = \left(x + 2\right) \left({x}^{3} - {2}^{3}\right)$

$\therefore = \left(x + 2\right) \left(x - 2\right) \left({x}^{2} + 2 x + {2}^{2}\right)$

$\therefore = \left(x + 2\right) \left(x - 2\right) \left({x}^{2} + 2 x + 4\right)$

Jun 27, 2018

Some are more straight forward to spot than others. Some you just have to try things out.

$\left(x + 2\right) \left(x - 2\right) \left({x}^{2} + 2 x + 4\right)$

#### Explanation:

Given: ${x}^{4} + 2 {x}^{3} - 8 x - 16$

First observation is that $2 \times 8 = 16$ so we have a possible connection there.

Lets have a look at the direct grouping:

$\left[\textcolor{w h i t e}{\frac{2}{2}} {x}^{4} + 2 {x}^{3} \textcolor{w h i t e}{\frac{2}{2}}\right] + \left[\textcolor{w h i t e}{\frac{2}{2}} - 8 x - 16 \textcolor{w h i t e}{\frac{2}{2}}\right]$

Consider the first brackets. If we factor out ${x}^{3}$ we end up with:

${x}^{3} \left[\textcolor{w h i t e}{\frac{2}{2}} x + 2 \textcolor{w h i t e}{\frac{2}{2}}\right] + \left[\textcolor{w h i t e}{\frac{2}{2}} - 8 x - 16 \textcolor{w h i t e}{\frac{2}{2}}\right]$

We can make the second brackets the same as the first if we factor out $\left(- 8\right)$ giving:

${x}^{3} \left[\textcolor{w h i t e}{\frac{2}{2}} x + 2 \textcolor{w h i t e}{\frac{2}{2}}\right] - 8 \left[\textcolor{w h i t e}{\frac{2}{2}} x + 2 \textcolor{w h i t e}{\frac{2}{2}}\right]$

We now factor out the $\left(x + 2\right)$ giving:

$\left(x + 2\right) \left({x}^{3} - 8\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Trying to factor the part "(x^3-8)

This is the same as $\left({x}^{3} - {2}^{3}\right)$

Can we manipulate a quadratic out of this? Lets have a play with:

$\left(x - 2\right) \left(\text{something}\right) = {x}^{3} - 8$

$\textcolor{b l u e}{\text{Dealing with the "x^3" term}}$

To obtain ${x}^{3}$ we try: (x-2)(x^2+?)=x^3-2x^2+?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{\text{Dealing with the "x^2" term}}$
but there is no $- 2 {x}^{2}$ term in ${x}^{3} - 8$ so we need to get rid of it.

Try: (x-2)(x^2+x+?)=x^3+x^2-2x^2-2x+? color(red)(" Fail")

Try: (x-2)(x^2+2x+?)=x^3+2x^2-2x^2-4x+? color(green)(" Works!")
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Dealing with the "-4x" term}}$

(x-2)(x^2+2x+?)=x^3-4x+?
but there is no $x$ term so we need to get rid of it.

Try: $\left(x - 2\right) \left({x}^{2} + 2 x + 2\right) = {x}^{3} - 4 x + 4 x \textcolor{red}{- 4} \leftarrow \textcolor{red}{\text{ Fail}}$

We got rid of the ${x}^{2}$ term but ended up with the wrong constant.

Try: $\left(x - 2\right) \left({x}^{2} + 2 x + 4\right) = {x}^{3} - 4 x + 4 x + 8 \leftarrow \textcolor{g r e e n}{\text{ Works}}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Putting it all back together}}$

$\left(x + 2\right) \left({x}^{3} - 8\right) \textcolor{w h i t e}{\text{d")->color(white)("d}} \left(x + 2\right) \left(x - 2\right) \left({x}^{2} + 2 x + 4\right)$

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$\textcolor{b l u e}{\text{Bottom line comment}}$

If you end up with $\left({a}^{3} - {b}^{3}\right)$ this becomes:

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

Where $\textcolor{w h i t e}{\text{d}} - \left(- b\right) a = + a b$