x to the 3rd-9x squared-8x-4/x+3=

\frac{x^3 -9x^2 - 8x-4}{x+3} = ?

Is that it?

Well here you go. I'll proceed with long division:
1. Divide x^3 by x and put the answer up there
2. Take the value you just obtained and multiply that by (x+3), putting the result below your numerator.
3. Subtract that result from your numerator.
4. Repeat with the next term.


iiiiiiiix^2 - 12x + 28
x + 3 | x^3 - 9x^2 - 8x - 4
x^3 + 3x^2
0 -12x^2 - 8x - 4
12x^2 -36x - 4
0 +28x - 4
28x + 84
0 - 88

So, your answer is x^2 + 12x + 28 with a remainder of -88.

I hope it helped! :)

Don't forget Unknown008 that with a remainder the answer is actually x^2+12x+28+(x+3)
(over -88)

(x+3)
(-88)

Actually, it's -\frac{88}{x+3} and not -\frac{x+3}{88}

Because:
x^3 - 9x^2 - 8x - 4 = (x+3)(x^2 -12x + 28) - 88

Dividing everything by (x+3) gives:
\frac{x^3 - 9x^2 - 8x - 4}{x+3} = (x^2 -12x + 28) - \frac{88}{x+3}

Sorry

Hey unky:

Why don't you teach mathwiz and us all how to write like that.

A sticky or a link.

You can show the tags using the noparse attribute.

e.g. This in bold is the Vbulletin code for This in bold

Maybe you can create a sticky someplace like "Forum Help" or the "math & Science section" Just a thought.

That's already there KISS, RickJ was the one who 'taught' me how to do so :). Check the sticky in the math-sciences page.