# How do you factor  2x^6-3x^4?

Jun 15, 2015

Factor out the greatest common factor ${x}^{4}$ to get ${x}^{4} \left(2 {x}^{2} - 3\right)$.

#### Explanation:

Factor out the GCF ${x}^{4}$.

$2 {x}^{6} - 3 {x}^{4}$ =

${x}^{4} \left(2 {x}^{2} - 3\right)$

Jun 15, 2015

$2 {x}^{6} - 3 {x}^{4} = {x}^{4} \left(2 {x}^{2} - 3\right) = {x}^{4} \left(\sqrt{2} x - \sqrt{3}\right) \left(\sqrt{2} x + \sqrt{3}\right)$

#### Explanation:

First separate out the common factor ${x}^{4}$ to get:

$2 {x}^{6} - 3 {x}^{4} = {x}^{4} \left(2 {x}^{2} - 3\right)$

The remaining quadratic factor can be treated as a difference of squares with irrational coefficients:

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

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

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

using the difference of squares identity:

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