# How do you factor y= 4t^5-12t^3+8t^2 ?

Dec 27, 2015

Separate out the common factor $4 {t}^{2}$, then use the coefficient zero sum check to find:

$4 {t}^{5} - 12 {t}^{3} + 8 {t}^{2} = 4 {t}^{2} \left(t - 1\right) \left(t - 1\right) \left(t + 2\right)$

#### Explanation:

First notice that all of the terms are divisible by $4 {t}^{2}$, so separate out that factor first:

$y = 4 {t}^{5} - 12 {t}^{3} + 8 {t}^{2}$

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

Next notice that the sum of the coefficients of the terms of ${t}^{3} - 3 t + 2$ is zero. That is, $1 - 3 + 2 = 0$. So $t = 1$ is a zero of this cubic and $\left(t - 1\right)$ is a factor:

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

Notice that the sum of the coefficients of the terms of $\left({t}^{2} + t - 2\right)$ is also zero, so there is another factor $\left(t - 1\right)$:

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

Putting this all together:

$4 {t}^{5} - 12 {t}^{3} + 8 {t}^{2} = 4 {t}^{2} \left(t - 1\right) \left(t - 1\right) \left(t + 2\right)$