# How do you factor  4x^5 - 9x^3?

Apr 24, 2017

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

#### Explanation:

Always take out a common factor first if there is one.

$4 {x}^{5} - 9 {x}^{3}$

$= {x}^{3} \left(4 {x}^{2} - 9\right) \text{ } \leftarrow$ difference of squares

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

To recognise Difference of squares:

• 2 terms
• minus sign
• perfect squares
• even indices

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

Apr 24, 2017

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

#### Explanation:

$4 {x}^{5} - 9 {x}^{3}$

Factor out ${x}^{3}$

${x}^{3} \left(4 {x}^{2} - 9\right)$

Here we realize that both $4 {x}^{2}$ and $9$ are perfect squares. So, we factor them as.

$\textcolor{red}{\left(\sqrt{A} + \sqrt{B}\right) \left(\sqrt{A} - \sqrt{B}\right)}$
I this case $\textcolor{red}{A} = 4 {x}^{2}$ and $\textcolor{red}{B} = 9$

So,
${x}^{3} \left(\sqrt{4 {x}^{2}} + \sqrt{9}\right) \left(\sqrt{4 {x}^{2}} - \sqrt{9}\right)$

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