Im stuck on this question:

$\displaystyle \frac{2x^2-x-6}{2x^2+3x-2}$ division $\displaystyle \frac{x^2-9}{x^2-x-6}$ X $\displaystyle \frac{4x^2-4x+1}{2x^2-5x+2}$

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- Mar 18th 2008, 02:49 PMeh501Simplifying radical expressions?
Im stuck on this question:

$\displaystyle \frac{2x^2-x-6}{2x^2+3x-2}$ division $\displaystyle \frac{x^2-9}{x^2-x-6}$ X $\displaystyle \frac{4x^2-4x+1}{2x^2-5x+2}$ - Mar 18th 2008, 02:54 PMtopher0805
Remember that when you are working with fractions, division can be expressed as such:

$\displaystyle \frac {a}{b}\div \frac {c}{d} = \frac {a}{b}\cdot \frac {d}{c}$

Now just multiply everything out, then factor out common terms. - Mar 18th 2008, 03:01 PMeh501
hmm..i still don't get what to do. Do i flip over the last fraction? And switch the X into a division sign?

- Mar 18th 2008, 05:26 PMJameson
Let $\displaystyle a=\frac{2x^2-x-6}{2x^2+3x-2}$, $\displaystyle b=\frac{x^2-9}{x^2-x-6}$, and $\displaystyle c=\frac{4x^2-4x+1}{2x^2-5x+2}$

I'm taking your problem to be $\displaystyle \frac{a}{b \times c}$ If so, you can use topher0805's suggestion and rewrite this as $\displaystyle a \times ( \frac{1}{b \times c})$ or you can simplify b x c first then work from there. Either way, you're just gonna have to do some algebra.

Can you get it started?