Division

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valnadam

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Feb 7, 2011
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33
Divide:
(x^2-1/x-3)/(x^2-x-2/3x-9)

I'm not good with division any help is appreciated
 
valnadam said:
Divide:
(x^2-1/x-3)/(x^2-x-2/3x-9)
Click to expand...
Are x-3 and 3x-9 also divisors?
If so then you MUST use extra bracketing:
[(x^2-1)/(x-3)] / [(x^2-x-2)/(3x-9)]

I suggest you make sure you understand WHY.
 
Re: Division check answer

valnadam said:
I get (3x-3/x-2)
Click to expand...
Before I answer, please repost that properly bracketed.
If you won't listen, neither will we.
 
valnadam said:
Divide:
[(x^2-1)/(x-3)] / [(x^2-x-2)/(3x-9)]

I get (3x-3/x-2)
Click to expand...

Here's what we are asking....

FIRST...post your problem with sufficient grouping symbols so that we don't need to be guessing at what you mean.

SECOND....show us how you have arrived at your answer. "I get....." tells us nothing about your procedures or where your mistakes might lie.

Ok?
 
valnadam said:
Divide:
[(x^2-1)/(x-3)] / [(x^2-x-2)/(3x-9)]

I get (3x-3/x-2)
Click to expand...

After all the complains about grouping symbols, what you wrote - as answer - translates to:

\(\displaystyle 3x - \frac{3}{x} - 2\)

If that's what you got - it is wrong.

Like Mrs? said - if you wanted us to show you where you went wrong - you'll need to show your work.
 
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