SOLUTION: I'm having a little trouble factoring trinomials. Ex. u^3= 14u^2 + 32u. I do: 0= -u^3 + 14u^2 + 32u. Then, according to my algebra book, I factor, or seem to, it to: 0= u(u^2 + 14u

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Question 911760: I'm having a little trouble factoring trinomials. Ex. u^3= 14u^2 + 32u. I do: 0= -u^3 + 14u^2 + 32u. Then, according to my algebra book, I factor, or seem to, it to: 0= u(u^2 + 14u + 32). Do I factor out the part in the parentheses? I can't seem to- paired factors of 32 that add up to 14... Or do I have it wrong? I just want to know the next step(s).
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39621) (Show Source):
You can put this solution on YOUR website!
You are stuck at this: and want to go further in factorization.

Now you are looking for two numbers so that their product is 32 and their sum is 14. Here is a list of combinations to check:

2 & 16;
4 & 8;
Those are all the integer combinations for that. Notice, 32=2*2*2*2*2.

Neither of those combinations work. Your factorization is finished.

Answer by MathTherapy(10555) (Show Source):
You can put this solution on YOUR website!

I'm having a little trouble factoring trinomials. Ex. u^3= 14u^2 + 32u. I do: 0= -u^3 + 14u^2 + 32u. Then, according to my algebra book, I factor, or seem to, it to: 0= u(u^2 + 14u + 32). Do I factor out the part in the parentheses? I can't seem to- paired factors of 32 that add up to 14... Or do I have it wrong? I just want to know the next step(s).

You have it all wrong!!
If the equation you described is: , then we get:

(u - 16)(u + 2) = 0
OR
You were correct, at first when you obtained: . From here on though, you
should've gotten: . This then results in: