# How do you factor 5x^4 + 2x^2 - 7?

Aug 11, 2018

$5 {x}^{4} + 2 {x}^{2} - 7 = \left(x + 1\right) \left(x - 1\right) \left(5 {x}^{2} + 7\right)$

#### Explanation:

$5 {x}^{4} + 2 {x}^{2} - 7 = 5 {x}^{4} + 7 {x}^{2} - 5 {x}^{2} - 7$
because,

$7 \times - 5 = - 35$

$7 + \left(- 5\right) = 2$

$= {x}^{2} \left(5 {x}^{2} + 7\right) - 1 \times \left(5 {x}^{2} + 7\right)$

$= \left({x}^{2} - 1\right) \times \left(5 {x}^{2} + 7\right)$

Further,

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

Thus,

$5 {x}^{4} + 2 {x}^{2} - 7 = \left(x + 1\right) \left(x - 1\right) \left(5 {x}^{2} + 7\right)$

Aug 11, 2018

$\left(x - 1\right) \left(x + 1\right) \left(5 {x}^{2} + 7\right)$

#### Explanation:

$\text{let } {x}^{2} = u$

$= 5 {u}^{2} + 2 u - 7$

$\text{factor the quadratic using the a-c method}$

$\text{the factors of the product } 5 \times - 7 = - 35$

$\text{which sum to "+2" are "-5" and } + 7$

$\text{use these factors to split the middle term}$

$5 {u}^{2} - 5 u + 7 u - 7 \leftarrow \textcolor{b l u e}{\text{factor by grouping}}$

$= \textcolor{red}{5 u} \left(u - 1\right) \textcolor{red}{+ 7} \left(u - 1\right)$

$\text{take out the "color(blue)"common factor } \left(u - 1\right)$

$= \left(u - 1\right) \left(\textcolor{red}{5 u + 7}\right)$

$\text{change u back to } {x}^{2}$

$= \left({x}^{2} - 1\right) \left(5 {x}^{2} + 7\right)$

$\left({x}^{2} - 1\right) \text{ is a "color(blue)"difference of squares}$

$= \left(x - 1\right) \left(x + 1\right) \left(5 {x}^{2} + 7\right)$