# How do you factor 8x^4+x ?

Jul 12, 2016

$x \left(2 x + 1\right) \left(4 {x}^{2} - 2 x + 1\right)$

#### Explanation:

List factors of both terms.

$8 {x}^{4} = 8 \cdot \textcolor{\mathmr{and} a n \ge}{x} \cdot x \cdot x \cdot x$

$x = \textcolor{\mathmr{and} a n \ge}{x}$

The common factors are just one $\textcolor{\mathmr{and} a n \ge}{x}$.
So let's pull out one $x$ from the expression.

$\textcolor{\mathmr{and} a n \ge}{x} \left(8 {x}^{3} + 1\right)$

We still need to factor $\left(8 {x}^{3} + 1\right)$. Take note that ${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$.

$x \left(2 x + 1\right) \left(4 {x}^{2} - 2 x + 1\right)$