Question

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

Answer 1

`y^3-12y^2+36y" = "y(y-6)^2`.

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.
`y^3-12y^2+36y" = "y(y-6)^2`.