Check out some similar questions!

rachelmorris · Nov 22, 2009, 10:37 PM

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

Unknown008 · Nov 23, 2009, 12:24 AM

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

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.

Code:

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! :)

mathwiz3502 · Nov 29, 2009, 06:37 AM

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

mathwiz3502 · Nov 29, 2009, 06:38 AM

(x+3)
(-88)

Unknown008 · Dec 1, 2009, 04:07 AM

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

mathwiz3502 · Dec 1, 2009, 03:31 PM

Sorry

KISS · Dec 1, 2009, 04:57 PM

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. [b]This in bold[/b] 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.

Unknown008 · Dec 1, 2009, 10:05 PM

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