# How do you factor completely: 10x^5 + 4x^4 + 8x^3?

Apr 21, 2018

$10 {x}^{5} + 4 {x}^{4} + 8 {x}^{3}$$= 2 {x}^{3} \left(5 {x}^{2} + 2 x + 4\right)$$= 10 {x}^{3} \left(x - \frac{1}{5} \left(- 1 + i \setminus \sqrt{19}\right)\right) \left(x - \frac{1}{5} \left(- 1 - i \setminus \sqrt{19}\right)\right)$

#### Explanation:

$10 {x}^{5} + 4 {x}^{4} + 8 {x}^{3}$

First we take out the obvious common factor of ${x}^{3}$. Let's take out the slightly less obvious factor of $2$ as well.

$= 2 {x}^{3} \left(5 {x}^{2} + 2 x + 4\right)$

That's pretty good. If we can factor the quadratic equation we can make more progress. To check if we can, we can check if the discriminant ${b}^{2} - 4 a c$ is a perfect square. Well ${2}^{2} - 4 \left(5\right) \left(4\right) = - 76$ is a negative number, which won't ever be a perfect square.

Depending on what grade we're in, we either stop here or factor using complex numbers. I'll continue.

Pro tip: The Shakespeare Quadratic Formula ($2 b$ or $- 2 b$) says ${x}^{2} - 2 b x + c$ has zeros $x = b \setminus \pm \sqrt{{b}^{2} - c}$ and $a {x}^{2} + 2 b x + c$ has zeros $x = \setminus \frac{1}{a} \left(- b \setminus \pm \sqrt{{b}^{2} - a c}\right)$.

$x = \frac{1}{5} \left(- 1 \setminus \pm \setminus \sqrt{- 19}\right) = \frac{1}{5} \left(- 1 \pm i \setminus \sqrt{19}\right)$

so we can factor

$5 {x}^{2} + 2 x + 4$

$= 5 \left(x - \frac{1}{5} \left(- 1 + i \setminus \sqrt{19}\right)\right) \left(x - \frac{1}{5} \left(- 1 - i \setminus \sqrt{19}\right)\right)$

and the entire expression

$10 {x}^{5} + 4 {x}^{4} + 8 {x}^{3}$

$= 10 {x}^{3} \left(x - \frac{1}{5} \left(- 1 + i \setminus \sqrt{19}\right)\right) \left(x - \frac{1}{5} \left(- 1 - i \setminus \sqrt{19}\right)\right)$