# How do you factor the expression 12t^8-75t^4?

Oct 17, 2016

$12 {t}^{8} - 75 {t}^{4} = 3 {t}^{4} \left(2 {t}^{2} - 5\right) \left(2 {t}^{2} + 5\right)$

$\textcolor{w h i t e}{12 {t}^{8} - 75 {t}^{4}} = 3 {t}^{4} \left(\sqrt{2} t - \sqrt{5}\right) \left(\sqrt{2} t + \sqrt{5}\right) \left(2 {t}^{2} + 5\right)$

#### Explanation:

$\textcolor{w h i t e}{}$
Difference of squares

The difference of squares identity can be written:

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

We will use this a couple of times.

$\textcolor{w h i t e}{}$
Factor $\boldsymbol{12 {t}^{8} - 75 {t}^{4}}$

First note that both terms are divisible by $3 {t}^{4}$, so separate that out as a factor first:

$12 {t}^{8} - 75 {t}^{4} = 3 {t}^{4} \left(4 {t}^{4} - 25\right)$

$\textcolor{w h i t e}{12 {t}^{8} - 75 {t}^{4}} = 3 {t}^{4} \left({\left(2 {t}^{2}\right)}^{2} - {5}^{2}\right)$

$\textcolor{w h i t e}{12 {t}^{8} - 75 {t}^{4}} = 3 {t}^{4} \left(2 {t}^{2} - 5\right) \left(2 {t}^{2} + 5\right)$

$\textcolor{w h i t e}{12 {t}^{8} - 75 {t}^{4}} = 3 {t}^{4} \left({\left(\sqrt{2} t\right)}^{2} - {\left(\sqrt{5}\right)}^{2}\right) \left(2 {t}^{2} + 5\right)$

$\textcolor{w h i t e}{12 {t}^{8} - 75 {t}^{4}} = 3 {t}^{4} \left(\sqrt{2} t - \sqrt{5}\right) \left(\sqrt{2} t + \sqrt{5}\right) \left(2 {t}^{2} + 5\right)$

The remaining quadratic $\left(2 {t}^{2} + 5\right)$ does not factor further with Real coefficients, since $2 {t}^{2} + 5 \ge 5 > 0$ for any Real value of $t$.