# [SOLVED] Factoring Polynomials

• Apr 16th 2009, 10:21 AM
giddy1945
[SOLVED] Factoring Polynomials
Hello, All,
I my teacher put our math test off till next class. I did not bring my notebook (thinking we were going to test), so I could not ask my 3 questions. She added the section I had questions about to the test. Please help me:

The directions are,"Factor each term"
I have three problems here, and if one of you guys could just work them out, I will be able to see the patterns. I know the answers, but I just need to see how they are worked out.

I tried googling "factoring polynomials) and this forum, but I guess I don't know how to look, so thanks again for your patience.

(x+3)(x+1)-y(x+1)

\$\displaystyle (3x-y)(x^2-2)+(x^2-2)\$

\$\displaystyle 3x(y+z)-9x(y+z)^2\$

Sorry 'bout the squares. There is a problem with my superscript.
• Apr 16th 2009, 10:31 AM
stapel
The polynomials you have posted are of the form AB + CB, where the common factor, B, can be pulled out front: AB + CB = B(A + C). To learn how this works, try here. (Wink)
• Apr 16th 2009, 10:58 AM
giddy1945
Most humble thanks, but I understand every bit of that link; however, these all have three sets of parentheses. The examples on that link only have two. These problems are more complex, and I am just failing to make the connection.
• Apr 16th 2009, 11:49 AM
Twig
hi
hi

You can try to look at it as if you want to "take it apart" as much as possible.
You look for common factors to pull out.

In your last, \$\displaystyle 3x(y+z) -9x(y+z)^{2} = 3x(y+z)(1 - 3(y+z)) \$
• Apr 16th 2009, 12:03 PM
giddy1945
Thank you. I will mark these solved. If a=(x+3), b=y, and x=(x+1) then i would have ax-bx. If I applied this principle to all three problems, they work out like majic. Thank you, guys.