# How do you factor y= 8t^4+ 32t ^3 + t + 4  ?

Dec 24, 2015

$y = \left(4 {t}^{2} - 2 t + 1\right) \left(2 t + 1\right) \left(t + 4\right)$

#### Explanation:

Factor by grouping. (Split into two groups of two.)

$y = \left(8 {t}^{4} + 32 {t}^{3}\right) + \left(t + 4\right)$

$y = 8 {t}^{3} \left(t + 4\right) + 1 \left(t + 4\right)$

Factor out a common $\left(t + 4\right)$.

$y = \left(8 {t}^{3} + 1\right) \left(t + 4\right)$

Recognize that $\left(8 {t}^{3} + 1\right)$ is a sum of cubes.

This is the following identity:

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

Thus, $a = 2 t , b = 1$, so

$8 {t}^{3} + 1 = \left(2 t + 1\right) \left(4 {t}^{2} - 2 t + 1\right)$

Replace this in the factored equation:

$y = \left(4 {t}^{2} - 2 t + 1\right) \left(2 t + 1\right) \left(t + 4\right)$