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

1 Answer
Dec 24, 2016

Answer:

#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#.