# How do you factor 36p^4-48p^3+16p^2?

Aug 25, 2016

$4 {p}^{2} \cdot {\left(3 p - 2\right)}^{2}$
OR
${\left(6 {p}^{2} - 4 p\right)}^{2}$

#### Explanation:

Fist we factor out ${p}^{2}$ leaving us with:
${p}^{2} \left(36 {p}^{2} - 48 p + 16\right)$
We can also factor out $4$ from all 3 terms so we get
$4 {p}^{2} \left(9 {p}^{2} - 12 p + 4\right)$
Now, looking at $9 {p}^{2} - 12 p + 4$ I see that $3 p \cdot 3 p = 9 {p}^{2}$ and $\left(- 2\right) \cdot \left(- 2\right) = 4$ and $\left(- 2 \cdot 3\right) + \left(- 2 \cdot 3\right) = - 12$. That means that:
$9 {p}^{2} - 12 p + 4 = \left(3 p - 2\right) \left(3 p - 2\right) = {\left(3 p - 2\right)}^{2}$

So final equation is $4 {p}^{2} \cdot {\left(3 p - 2\right)}^{2}$

Another answer is if we recognize that $4 {p}^{2} = {\left(2 p\right)}^{2}$ and distribute $2 p$ into each of the $\left(3 p - 2\right)$ leaving each term as $\left(6 {p}^{2} - 4 p\right)$ so the final answer would be:
${\left(6 {p}^{2} - 4 p\right)}^{2}$