# Quick Help Dividing Polynomials

1. Oct 14, 2007

### marcflores

1. The problem statement, all variables and given/known data

$$(6x^4-3x^2+x-4) / (2x^2+1)$$

2. Relevant equations

Relevant equations???

3. The attempt at a solution

Here is my attempt, but I want to make sure that I didn't break any laws by changing the number to be divided by switching the last two terms around by using the commutative law of addition:

$$[6x^4-3x^2+(-4)+x]$$ -- is this okay so far?

So, I wrote $$2x^2+1$$ divided by $$[6x^4-3x^2+(-4)+x]$$

Then I wrote it out like long division (can't find the division symbol in laTex) and here are the steps I took:

$$(3x^2)(2x^2+1) = 6x^4+3x^2$$

Then, I subtracted the product from the first two terms of the dividend:

$$(6x^4-3x^2) - (6x^4+3x^2) = -6x^2$$

Carry down the +(-4) from the dividend and I have:

$$-6x^2-4$$

Then I multiply the divisor by negative three and subtract the product from $$-6x^2-4$$:

$$(2x^2+1)(-3) = -6x^2-3$$

That leaves me with -1 and I carry down the x from the dividend leaving:

-1 + x

So, my answer is $$3x^2-3+\frac{-1+x}{2x^2+1}$$

I just want to know if this is correct or if I was wrong in redistributing the terms in the original dividend. Thanks.

2. Oct 14, 2007

### marcflores

Jeez, re-reading my post even left me confused. It's too bad I can't replicate it on the forums as I have it written on my paper.

3. Oct 14, 2007

### bob1182006

Usually you want to keep the terms in decreasing power so you know for sure you do them in that order. you don't want to do 3x^2 before doing 6x^4 here.
You can change the terms around just be sure to do them in the right order otherwise you could get mixed up and get the answer wrong.

4. Oct 14, 2007

### marcflores

I'll keep that in mind. Thanks for the input and help.

I think the reason I did 3x^2 was to find a monomial that would produce the product 6x^4 when multiplied by 2x^2 (the first term in the divisor).