# How do you factor 8x^6 + 7x^3 - 1?

Sep 1, 2016

$\left(1 {x}^{3} + 1\right) \left(8 {x}^{3} - 1\right)$

#### Explanation:

This is in the form of a quadratic because.

$8 {\left({x}^{3}\right)}^{2} + 7 {\left({x}^{3}\right)}^{1} - 1$

Find the factors of 8 (and 1) which subtract to give 7

$8 - 1 = 7$

The signs will be different (because of -1)
There must be more positives (because of +7)

The factors of -1 can only be +1 and -1

(? x^3 +1)(? x^3-1)

Place the 8 and the 1 so that we will get +8 and -1. when we multiply.

$\left(1 {x}^{3} + 1\right) \left(8 {x}^{3} - 1\right)$

Set $t = {x}^{3}$ then the given equation is transformed to

$8 {t}^{2} + 7 t - 1$

$8 {t}^{2} + \left(8 - 1\right) t - 1$

$8 {t}^{2} + 8 t - t - 1$

$8 t \left(t + 1\right) - \left(t + 1\right)$

$\left(t + 1\right) \cdot \left(8 \cdot t - 1\right)$

Reverse to x we have that

$\left(1 + {x}^{3}\right) \cdot \left(8 \cdot {x}^{3} - 1\right)$

Now ${x}^{3} + 1$ can be factored further as follows

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

and $8 \cdot {x}^{3} - 1$ can be factored as follows

$8 {x}^{3} - 1 = \left(2 x - 1\right) \left(4 {x}^{2} + 2 x + 1\right)$

Hence finally we get

$8 {x}^{6} + 7 {x}^{3} - 1 = \left(x + 1\right) \cdot \left({x}^{2} - x + 1\right) \cdot \left(2 \cdot x - 1\right) \cdot \left(4 \cdot {x}^{2} + 2 x + 1\right)$