# Factoring polynomials

• Mar 10th 2011, 06:51 PM
Enkir
Factoring polynomials
Please forgive me for any placement errors of the thread or ah..pretty much any errors here. Being new and all I feel a bit intimidated here.

Anyway, as the topic says I'm currently stuck with factoring some polynomials. We're given one factor and we have to find the rest. Process is through synthetic substitution.

$3x^³ - 4x^² - 17x + 6; x + 2$
Now, I can work through most problems dealing with this. But what I'm stuck on...is the very first number. Does it effect the outcome of the problem at all? Anything you should do with it before attempting to solve or do you just go through with it and mess with it at the end?
• Mar 10th 2011, 07:01 PM
TheChaz
When given a factor (such as x + 2), set it equal to zero.
Then we will have x = -2. This is the number that goes "in the box" when doing synthetic division.

There's no way I'm going to try to visually reproduce the process (!), so I'll just talk you through it...
Your coefficients are 3, -4, -17, and 6.

Bring down the
3. Then 3*-2 + -4 =
-10. Then -10*-2 + -17 =
3. Then 3*-2 + 6 = 0 (as it should).

You now have coefficients 3, -10, and 3, corresponding to the quadratic

$3x^2 - 10x + 3$

Now you can factor this into (3x - 1)(x - 3), but don't lose the original factor!

Your full factorization is (3x - 1)(x - 3)(x + 2)
• Mar 10th 2011, 07:08 PM
Mahanen
You can just work through it normally:

Using synthetic division: h=-2 as $(x+2)$ is a factor (write the coefficients, drop the first, multiply up by 'h' and then add down the next column etc)

3 -4 -17 6
+ -6 -20 6
3 -10 3 0

This will leave you with $3x^2-10x+3$ as the result of the first factorization and then factoring that can be done by breaking that into $3x^2 -9x-1x+3 = 3x(x-3)-1(x-3) = (3x-1)(x-3)$

This gives you the three factors of $(x+2)(3x-1)(x-3)$
• Mar 10th 2011, 07:20 PM
TheChaz
Quote:

Originally Posted by Mahanen
...

3 -4 -17 6
+ -6 -20 6
3 -10 3 0
...

I guess that wasn't so bad!