# How do you factor  x^4 + 7x^3 - 8x - 56 = 0 ?

Sep 15, 2016

${x}^{4} + 7 {x}^{3} - 8 x - 56 = \left(x - 2\right) \left({x}^{2} + 2 x + 4\right) \left(x + 7\right)$

$\textcolor{w h i t e}{{x}^{4} + 7 {x}^{3} - 8 x - 56} = \left(x - 2\right) \left(x + 1 - \sqrt{3} i\right) \left(x + 1 + \sqrt{3} i\right) \left(x + 7\right)$

with corresponding zeros $x = 2$, $x = - 7$ and $x = - 1 \pm \sqrt{3} i$

#### Explanation:

The difference of cubes identity can be written:

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

We will use this with $a = x$ and $b = 2$ later.

$\textcolor{w h i t e}{}$
Note that the ratio between the first and second terms is the same as that between the third and fourth terms. So this quartic will factor by grouping:

${x}^{4} + 7 {x}^{3} - 8 x - 56 = \left({x}^{4} + 7 {x}^{3}\right) - \left(8 x + 56\right)$

$\textcolor{w h i t e}{{x}^{4} + 7 {x}^{3} - 8 x - 56} = {x}^{3} \left(x + 7\right) - 8 \left(x + 7\right)$

$\textcolor{w h i t e}{{x}^{4} + 7 {x}^{3} - 8 x - 56} = \left({x}^{3} - 8\right) \left(x + 7\right)$

$\textcolor{w h i t e}{{x}^{4} + 7 {x}^{3} - 8 x - 56} = \left({x}^{3} - {2}^{3}\right) \left(x + 7\right)$

$\textcolor{w h i t e}{{x}^{4} + 7 {x}^{3} - 8 x - 56} = \left(x - 2\right) \left({x}^{2} + 2 x + {2}^{2}\right) \left(x + 7\right)$

$\textcolor{w h i t e}{{x}^{4} + 7 {x}^{3} - 8 x - 56} = \left(x - 2\right) \left({x}^{2} + 2 x + 4\right) \left(x + 7\right)$

The Real zeros corresponding to the linear factors we have found are $x = 2$ and $x = - 7$

The remaining quadratic factor has no Real zeros, so will only factor using Complex coefficients.

${x}^{2} + 2 x + 4 = {x}^{2} + 2 x + 1 + 3$

$\textcolor{w h i t e}{{x}^{2} + 2 x + 4} = {\left(x + 1\right)}^{2} - {\left(\sqrt{3} i\right)}^{2}$

$\textcolor{w h i t e}{{x}^{2} + 2 x + 4} = \left(\left(x + 1\right) - \sqrt{3} i\right) \left(\left(x + 1\right) + \sqrt{3} i\right)$

$\textcolor{w h i t e}{{x}^{2} + 2 x + 4} = \left(x + 1 - \sqrt{3} i\right) \left(x + 1 + \sqrt{3} i\right)$