Polynomial factoring however....  TOPIC_SOLVED

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Polynomial factoring however....

Postby mark.ump on Mon Jul 11, 2011 9:02 pm

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. Image 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
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Postby stapel_eliz on Mon Jul 11, 2011 9:06 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! :wink:
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Re: Polynomial factoring however....

Postby mark.ump on Mon Jul 11, 2011 11:06 pm

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)
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Postby stapel_eliz on Tue Jul 12, 2011 3:38 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:

. . . . .

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