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

##### 1 Answer

#### Answer:

#### Explanation:

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

Here, all three terms have a

#y# in common.

#y^3-12y^2+36y#

#=y(y^2-12y+36)#

(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^2-12y+36# , because we can find two numbers that add to#"-12"# and multiply to#36# —namely,#"-6"# and#"-6"# .

#y(y^2-12y+36)#

#=y(y-6)(y-6)#

(Once again, if we re-multiply#(y-6)(y-6)# , we'd end up with#y^2-6y-6y+36# , or#y^2-12y+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#(y-6)# factors into a single one, with a power:

#y(y-6)(y-6)#

#=y(y-6)^2#

And there we have it.