Algebraic Division Help

[Josh M.]

I didn't really know where to put this, but I thought I'd post it in General Math. Any way, here's the problem, I'd like it solved step by step, please.

a4 (a to the fourth power) + Oa3 (a to the third power) + 9a2 (a squared) + Oa

divided by:

a2 (a squared) - 3a + 9

Thanks in advance!

Josh

[matt grime]

what is O?

[kreil]

Weird, you have the same name as me (mine is josh meyer). Anyway, heres the problem:

\frac{a^4+9a^2}{a^2-3a+9}

So we set up long division as you did, with 0's in for all the non-existent powers of x. It's hard to write this out on the site, so I am doing it on paper and describing it step by step:

1. a^2 goes in to a^4 a^2 times, so write a^2 above the 9a^2

2. Multiply the a^2-3a+9 term by that a^2 and write all the terms obtained underneath their proper powers (cubics under cubics etc). Then change the signs on all these terms and subtract everything. (you should get 0a^4+3a^3+0a^2)

3. a^2 goes into 3a^3 3a times, so write this 3a above the 0a, multiply the a^2-3a+9 term by it, change the signs on these terms, and subtract. You ought to get
-9a^2-27a.

4. a^2 goes into -9a^2 -9 times, so write this above, multiply everything out, switch signs, and subtract.

so, the answer is a^2+3a-9 with a remainder of 81.