1. Nov 3, 2009

### acuraintegra9

Divide the polynomials by using long division.
(-9x^6+7x^4-2x^3+5)/(3x^4-2x+1)

When I attempted it I started by pulling using 3x^2 . multiplied that by the (3x^4-2x+1) and from there I had to use a fraction of 7/3 or something and then couldnt divide into x cubed.

If anyone can help me that would help out so much.. All the examples in my book, that we talked about in class, and the ones online are mostly simple ones that work out evenly, are not that high of exponents etc.

2. Nov 3, 2009

### symbolipoint

Use the polynomials in their expanded forms. Your first partial quotient should be -3x^2. Multiply this by the divisor and subtract the result from the dividend to obtain your revised dividend in order to continue the process.

(note carefully, "-3x^2" means "negative three times x squared")

Last edited: Nov 3, 2009
3. Nov 3, 2009

### acuraintegra9

I did that much, but then the tricky part is when I add it in whats left after eliminating the x6 and x5 I get .. 7x^4-8x^3+3x^2+0x +5

dont know what to do from there.. do I have to multiply the 3x^4 by some sort of fraction to get it to equal to 7 to cancel out..

4. Nov 4, 2009

### acuraintegra9

please help! not sure what to do .. I get stuck at the next step .. not sure what to multiply by, dont cancel the x to the fourth's out

5. Nov 4, 2009

### symbolipoint

The second partial division and subtraction yielded a remainder. The process needed inclusion of the last both two terms in order to perform a sensible second subtraction; I needed to use all FIVE terms of the divisor, which is why the last two expanded form terms of the dividend were needed.

I have no easy way to typeset or express the numeric process in this forum. Let me just say, my final result after simplification was
-3x^2+(7/3)+(-8x^3 + 3x^2-(14/3)x+4)/(3x^4 -2x+1)

See if you can duplicate that! So many symbols and slightly lengthy, maybe I made an error.

6. Nov 4, 2009

### acuraintegra9

thats pretty much what I came up with , not sure though with it all being a remainder like that.. thats half the reason I thought my answer was wrong

7. Nov 4, 2009

### symbolipoint

What course are you in? How does your textbook treat this topic? Did your teacher mention or give any example like the one you posted?

You could try checking my result using multiplying the result by the divisor; it should be found equivalent to the dividend.