How do you factor completely #y^3 - 12y^2 + 36y#?
1 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.
