Dividing a⁴ + 9a² by a² - 3a + 9

[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, here's 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 into $$a^4$$ $$a^2$$ 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.