# How do you solve using factoring: 4x^4 + 23x^2 = 72?

Oct 29, 2016

$x = \pm \frac{3}{2} \text{ AND } x = \pm 2 \sqrt{2} i$

#### Explanation:

Rewrite in standard form, $a {x}^{2} + b x + c = 0$.

$\to 4 {x}^{4} + 23 {x}^{2} - 72 = 0$

This is a fourth degree trinomial, so we can factor it using trinomial factoring rules, such as in $a {x}^{2} + b x + c , a \ne 0 , 1$, to factor you need to find two numbers that multiply to $a c$ and that add to $b$.

$4 {x}^{4} + 32 {x}^{2} - 9 {x}^{2} - 72 = 0$

$4 {x}^{2} \left({x}^{2} + 8\right) - 9 \left({x}^{2} + 8\right) = 0$

$\left(4 {x}^{2} - 9\right) \left({x}^{2} + 8\right) = 0$

${x}^{2} = \frac{9}{4} \text{ AND } {x}^{2} = - 8$

$x = \pm \frac{3}{2} \text{ AND } x = \pm 2 \sqrt{2} i$

Hopefully this helps!