# Thread: Factoring help

1. ## 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!

2. 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

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

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

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

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

4. 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

5. $px^3+(p-q)x^2+(-2p-q)x+2q = -(x-1)(x+2)(q-px)$ 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.

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

7. 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!

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

9. 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.

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