## Polynomial factoring however....

Simplificatation, evaluation, linear equations, linear graphs, linear inequalities, basic word problems, etc.

### Polynomial factoring however....

The however part deals with the fact that the certain book that I'm using for self-study throws a curve ball so to say. The chapter of solving polynomials doesn't explain how to solve this particular problem. I need a good explanation on why the problems 87 and 88 are solved completely different from the others.
Here is the link to the problem because the photo won't show 88 on this site too big of a photo I guess. http://tinypic.com/view.php?pic=25ak94g&s=7
mark.ump

Posts: 3
Joined: Mon Jul 11, 2011 8:55 pm

What was left when you factored in pairs and took the common factor out front? In other words, after you'd done the exact same initial steps, did you end up with the exact same final results, or did you get something that could be further factored?

It would probably help if you showed your work. Thank you!

stapel_eliz

Posts: 1802
Joined: Mon Dec 08, 2008 4:22 pm

### Re: Polynomial factoring however....

I first saw that a b and x are common factors in the problem. At first I knew that a and b needed to come out. I just didn't know how to do it. I thought ab but then I saw that a and b aren't ever combined together with multiplication. I figured it was a-b or a+b. I'm not sure why it has to be a+b but it does. Then I had (a+b)(x^2+x^2-4-4) which equaled to me (a+b)(2x^2-8)
mark.ump

Posts: 3
Joined: Mon Jul 11, 2011 8:55 pm

mark.ump wrote:...Then I had (a+b)(x^2+x^2-4-4) which equaled to me (a+b)(2x^2-8)

Now take the common factor of 2 out front, leaving you with:

. . . . .$2(a\, +\, b)(x^2\, -\, 4)$

All that is left is factoring the difference of squares to complete the exercise.

stapel_eliz

Posts: 1802
Joined: Mon Dec 08, 2008 4:22 pm