# How do you factor completely y^3 - 12y^2 + 36y?

Dec 24, 2016

${y}^{3} - 12 {y}^{2} + 36 y \text{ = } y {\left(y - 6\right)}^{2}$.

#### Explanation:

Step 1: Find anything common across all terms; factor this out.

Here, all three terms have a $y$ in common.

${y}^{3} - 12 {y}^{2} + 36 y$

$= y \left({y}^{2} - 12 y + 36\right)$

(Notice if we were to re-distribute this $y$, we'd end up with the starting polynomial.)

Step 2: Do any trinomial factoring (or "division") that's possible.

Here, we can factor ${y}^{2} - 12 y + 36$, because we can find two numbers that add to $\text{-12}$ and multiply to $36$—namely, $\text{-6}$ and $\text{-6}$.

$y \left({y}^{2} - 12 y + 36\right)$

$= y \left(y - 6\right) \left(y - 6\right)$

(Once again, if we re-multiply $\left(y - 6\right) \left(y - 6\right)$, we'd end up with ${y}^{2} - 6 y - 6 y + 36$, or ${y}^{2} - 12 y + 36$, which is the trinomial we started with.)

Step 3: Repeat Step 2 if possible.

In our case, this is as far as we can go; all factors have $y$ to the smallest powers we can get. The only other thing we can do is "merge" the two $\left(y - 6\right)$ factors into a single one, with a power:

$y \left(y - 6\right) \left(y - 6\right)$

$= y {\left(y - 6\right)}^{2}$

And there we have it.
${y}^{3} - 12 {y}^{2} + 36 y \text{ = } y {\left(y - 6\right)}^{2}$.