SOLUTION: Factoring Polynomials. I had a horrible Algebra II teacher last year and learned almost nothing that i was supposed to because of it. I dont think I am very good at factoring. Th

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Factoring Polynomials. I had a horrible Algebra II teacher last year and learned almost nothing that i was supposed to because of it. I dont think I am very good at factoring. Th      Log On


   

Question 24564: Factoring Polynomials. I had a horrible Algebra II teacher last year and learned almost nothing that i was supposed to because of it. I dont think I am very good at factoring. This is for College Algebra. I am dually enrolled in high school and a community college. The directions say to factor out, relative to the integers, all factors common to all terms. The problem is..
2w(y-2z)-y(y-2Z) I went ahead and simplified the problem, believing that to be the right thing to do. I got..2wy-4wz-y^2+2yz The answer is..(2w-x)(y-2z)
Thank you so much. I absolutely must learn this stuff. Any correct explanation will be taken to heart.
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Well, your answer is almost correct...except for a small typo. You have an x in the answer when there is no x in the original expression. Here are the steps:
Factor:
2w%28y-2z%29-y%28y-2z%29 It is not neccesary to simplify as you did because, if you look at the expression, you will see that each term has a common factor of (y-2z) and this, of course, can be factored out.
%28y-2z%29%282w-y%29 ...and there you have it!