# How do you factor 4x^4-5x^2-9?

May 26, 2018

$\left(2 x - 3\right) \left(2 x + 3\right) \left(x + i\right) \left(x - i\right)$

#### Explanation:

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

$\text{the factors of the product } 4 \times - 9 = - 36$

$\text{which sum to - 5 are + 4 and - 9}$

$\text{split the middle term using these factors}$

$4 {x}^{4} + 4 {x}^{2} - 9 {x}^{2} - 9 \leftarrow \textcolor{b l u e}{\text{factor by grouping}}$

$= 4 {x}^{2} \left({x}^{2} + 1\right) - 9 \left({x}^{2} + 1\right)$

$\text{take out the "color(blue)"common factor } \left({x}^{2} + 1\right)$

$= \left({x}^{2} + 1\right) \left(4 {x}^{2} - 9\right)$

$4 {x}^{2} - 9 \text{ is a "color(blue)"difference of squares}$

$\text{which factors in general as}$

•color(white)(x)a^2-b^2=(a-b)(a+b)

$4 {x}^{2} = {\left(2 x\right)}^{2} \Rightarrow a = 2 x$

$9 = {\left(3\right)}^{2} \Rightarrow b = 3$

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

${x}^{2} + 1 \text{ can also be factored using difference of squares}$

$= \left(x + i\right) \left(x - i\right)$

$4 {x}^{4} - 5 {x}^{2} - 9 = \left(2 x - 3\right) \left(2 x + 3\right) \left(x + i\right) \left(x - i\right)$