SOLUTION: Hi, I have a question! If (x-1)(x^2-4)+2(x-1)(x+2)=(x-1)P, then P= (A) x^2-2 (B) x^2 (C) x(x+2) (D) x^2+2 (E) (x+2)^2 I know that the answer is C, but I am confused at

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Hi, I have a question! If (x-1)(x^2-4)+2(x-1)(x+2)=(x-1)P, then P= (A) x^2-2 (B) x^2 (C) x(x+2) (D) x^2+2 (E) (x+2)^2 I know that the answer is C, but I am confused at      Log On


   

Question 604723: Hi, I have a question!
If (x-1)(x^2-4)+2(x-1)(x+2)=(x-1)P, then P=
(A) x^2-2
(B) x^2
(C) x(x+2)
(D) x^2+2
(E) (x+2)^2
I know that the answer is C, but I am confused at a certain step...
So I got to:
(x+2)(x-2)+2(x+2)=P
but how do I go from that to:
x(x+2)
Thanks!
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
If:
%28x-1%29%28x%5E2-4%29%2B2%28x-1%29%28x%2B2%29+=+%28x-1%29P Solve for P: First, factor (x-1) on the left side.
%28x-1%29%28%28x%5E2-4%29%2B2%28x%2B2%29%29+=+%28x-1%29P Divide both sides by (x-1)
%28x%5E2-4%29%2B2%28x%2B2%29+=+P Simplify.
x%5E2-4%2B2x%2B4+=+P and so...
P+=+x%5E2%2B2x Factor an x.
P+=+x%28x%2B2%29