Factoring help

**starswept** · July 7th 2007, 01:39 PM

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!

**topsquark** · July 7th 2007, 01:51 PM

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

**janvdl** · July 7th 2007, 01:52 PM

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

$px^3 + px^2 -2px - qx^2 - qx + 2q$

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

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

$(px - q)(x + 2)(x - 1)$

**janvdl** · July 7th 2007, 01:52 PM

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

**rualin** · July 7th 2007, 02:03 PM

$px^3+(p-q)x^2+(-2p-q)x+2q = -(x-1)(x+2)(q-px)$ says QuickMath

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

**rualin** · July 7th 2007, 02:09 PM

You factored wrong on $px(x^2 + x - 2) - q(x^2 - x + 2)$ , Janvdl. The rest looks fine.

**Jhevon** · July 7th 2007, 02:10 PM

Originally Posted by rualin

$px^3+(p-q)x^2+(-2p-q)x+2q = -(x-1)(x+2)(q-px)$ says QuickMath

We can make a list of the possible zeros, which include $\pm 1, \pm 2, \pm q$ 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!

**rualin** · July 7th 2007, 02:22 PM

I also prefer that approach but when it seems difficult to factor, I find it better for me to use the factor theorem.

**janvdl** · July 7th 2007, 02:27 PM

Originally Posted by Jhevon

..Good Job Janvdl. You are the best mathematician in this post!

I'm flattered.

Couldnt have done it without you and Rualin who showed me that mistake though. Thanks guys.

**Soroban** · July 7th 2007, 02:48 PM

Hello, starswept!

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

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

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

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

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

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

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

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