Hi, I need help with this question:

Factor the expressions px^3 + (p-q)x^2 + (-2p-q)x + 2q and abx^3 + (a-2b-ab)x^2 + (2b-a-2)x + 2

I've tried a few different ways, but none of them work out. Thank you for any help!

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- July 7th 2007, 01:39 PMstarsweptFactoring help
Hi, I need help with this question:

Factor the expressions px^3 + (p-q)x^2 + (-2p-q)x + 2q and abx^3 + (a-2b-ab)x^2 + (2b-a-2)x + 2

I've tried a few different ways, but none of them work out. Thank you for any help! - July 7th 2007, 01:51 PMtopsquark
- July 7th 2007, 01:52 PMjanvdlQuote:

- July 7th 2007, 01:52 PMjanvdl
- July 7th 2007, 02:03 PMrualin
says QuickMath

We can make a list of the possible zeros, which include and check which yield a zero either through substitution or synthetic division. - July 7th 2007, 02:09 PMrualin
You factored wrong on , Janvdl. The rest looks fine.

- July 7th 2007, 02:10 PMJhevon
i was just about to make that suggestion

i noticed almost immediately that x - 1 was a factor by the factor theorem. performing long division twice will get you the answer you have

coincidentally, (x - 1) is a factor of the second one as well

janvdl's method is still easiest by far though. just make sure you factor correctly. factoring out a -q from the second set changes all the signs inside the bracket

EDIT: Good Job Janvdl. You are the best mathematician in this post! - July 7th 2007, 02:22 PMrualin
I also prefer that approach but when it seems difficult to factor, I find it better for me to use the factor theorem.

- July 7th 2007, 02:27 PMjanvdl
- July 7th 2007, 02:48 PMSoroban
Hello, starswept!

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