# How do you factor completely: 8x^4-28x^6?

Jul 28, 2015

$8 {x}^{4} - 28 {x}^{6} = 4 {x}^{4} \left(2 - 7 {x}^{2}\right) = 4 {x}^{4} \left(\sqrt{2} - \sqrt{7} x\right) \left(\sqrt{2} + \sqrt{7} x\right)$

#### Explanation:

Both terms are divisible by $4 {x}^{4}$ so we can separate out that factor first...

$8 {x}^{4} - 28 {x}^{6} = 4 {x}^{4} \left(2 - 7 {x}^{2}\right)$

Next, if we allow irrational coefficients, $2 - 7 {x}^{2}$ can be treated as a difference of squares, so we can use the difference of squares identity:

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

with $a = \sqrt{2}$ and $b = \sqrt{7} x$

So:

$2 - 7 {x}^{2} = {\left(\sqrt{2}\right)}^{2} - {\left(\sqrt{7} x\right)}^{2} = \left(\sqrt{2} - \sqrt{7} x\right) \left(\sqrt{2} + \sqrt{7} x\right)$

Putting it all together:

$8 {x}^{4} - 28 {x}^{6} = 4 {x}^{4} \left(2 - 7 {x}^{2}\right) = 4 {x}^{4} \left(\sqrt{2} - \sqrt{7} x\right) \left(\sqrt{2} + \sqrt{7} x\right)$