# Factoring help

• Jul 7th 2007, 01:39 PM
starswept
Factoring 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!
• Jul 7th 2007, 01:51 PM
topsquark
Quote:

Originally Posted by starswept
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!

There's either typos in this or there's something about the problem you aren't telling us. I don't believe these can be factored in general.

-Dan
• Jul 7th 2007, 01:52 PM
janvdl
Quote:

$\displaystyle px^3 + (p-q)x^2 + (-2p-q)x + 2q$
$\displaystyle px^3 + px^2 -2px - qx^2 - qx + 2q$

$\displaystyle px(x^2 + x - 2) - q(x^2 - x - 2)$

$\displaystyle px(x^2 + x - 2) - q(x^2 - x - 2)$

$\displaystyle (px - q)(x + 2)(x - 1)$
• Jul 7th 2007, 01:52 PM
janvdl
Quote:

Originally Posted by topsquark
There's either typos in this or there's something about the problem you aren't telling us. I don't believe these can be factored in general.

-Dan

Oh yes they can :D
• Jul 7th 2007, 02:03 PM
rualin
$\displaystyle px^3+(p-q)x^2+(-2p-q)x+2q = -(x-1)(x+2)(q-px)$ says QuickMath

We can make a list of the possible zeros, which include $\displaystyle \pm 1, \pm 2, \pm q$ and check which yield a zero either through substitution or synthetic division.
• Jul 7th 2007, 02:09 PM
rualin
You factored wrong on $\displaystyle px(x^2 + x - 2) - q(x^2 - x + 2)$, Janvdl. The rest looks fine.
• Jul 7th 2007, 02:10 PM
Jhevon
Quote:

Originally Posted by rualin
$\displaystyle px^3+(p-q)x^2+(-2p-q)x+2q = -(x-1)(x+2)(q-px)$ says QuickMath

We can make a list of the possible zeros, which include $\displaystyle \pm 1, \pm 2, \pm q$ and check which yield a zero either through substitution or synthetic division.

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!
• Jul 7th 2007, 02:22 PM
rualin
I also prefer that approach but when it seems difficult to factor, I find it better for me to use the factor theorem.
• Jul 7th 2007, 02:27 PM
janvdl
Quote:

Originally Posted by Jhevon
..Good Job Janvdl. You are the best mathematician in this post!

I'm flattered. :D Couldnt have done it without you and Rualin who showed me that mistake though. Thanks guys. :)
• Jul 7th 2007, 02:48 PM
Soroban
Hello, starswept!

Quote:

$\displaystyle abx^3 + (a-2b-ab)x^2 + (2b-a-2)x + 2$

Expand: .$\displaystyle abx^3 + ax^2 - 2bx^2 - abx^2 + 2bx - ax - 2x + 2$

Re-arrange: .$\displaystyle \underbrace{abx^3 - abx^2} + \underbrace{ax^2 - ax} - \underbrace{2bx^2 + 2bx} - \underbrace{2x + 2}$

. . Factor: .$\displaystyle abx^2(x-1) + ax(x-1) - 2bx(x-1) - 2(x-1)$

. . Factor: .$\displaystyle (x-1)\,(\underbrace{abx^2 + ax} - \underbrace{2bx - 2})$

. . Factor: .$\displaystyle (x-1)\,[ax(bx +1) - 2(bx+1)]$

. . Factor: .$\displaystyle (x-1)\,(ax-2)\,(bx+1)$