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Division
- Thread starter valnadam
- Start date
Are x-3 and 3x-9 also divisors?valnadam said:Divide:
(x^2-1/x-3)/(x^2-x-2/3x-9)
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
If you won't listen, neither will we.
Before I answer, please repost that properly bracketed.valnadam said:I get (3x-3/x-2)
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)
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?
D
Deleted member 4993
Guest
valnadam said:Divide:
[(x^2-1)/(x-3)] / [(x^2-x-2)/(3x-9)]
I get (3x-3/x-2)
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.