Finding Solutions (Polynomial 2)

[adr8]

I just have a quick question. I have to find the solutions. I already got the answers for the following problem:

(x^2-4x+4)² (x²+6x+5)³ I factored them out which came out to [(x-2)(x-2)]²[(x+5)(x+1)]³. So my solutions were 2, -5, and -1. My only question is the exponents such as the 2 and the 3, did I have to use them? Or was it fine like the way how I solved the problem?

 

[adr8]

WHY are ALL your questions "quick"? :cool:

Can you please post the ORIGINAL problem IN FULL?
What solutions are you trying to find? You are giving us no equations.
Do you intend this : (x^2-4x+4)² (x²+6x+5)³ = 0 ?
ONLY equations can be solved; equations ALL have an equal sign.

Oh its because I think I got it lol, but I just want to know if I did them right since there is something different in the problem.

Just the solutions to this equation. The original problem is the one that I showed u: (x^2-4x+4)² (x²+6x+5)³ = 0


I factored them out (x-2)(x-2) (x+1)(x+5), and I set them each equal to zero. My answers were x=2, -1, and -5. The only thing that gets me wondering is the ³ and ². I mean do I have to use them or is it okay the way how I solved it?

 

[adr8]

My bad I forgot to add the =0. But do I have to do something with the ²and³ ? Or was it fine the way how I solved it?