Simplifying radical expressions?

**eh501** · March 18th 2008, 02:49 PM

Im stuck on this question:

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

**topher0805** · March 18th 2008, 02:54 PM

Remember that when you are working with fractions, division can be expressed as such:

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

Now just multiply everything out, then factor out common terms.

**eh501** · March 18th 2008, 03:01 PM

hmm..i still don't get what to do. Do i flip over the last fraction? And switch the X into a division sign?

**Jameson** · March 18th 2008, 05:26 PM

Originally Posted by eh501

Im stuck on this question:

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

Let $a=\frac{2x^2-x-6}{2x^2+3x-2}$ , $b=\frac{x^2-9}{x^2-x-6}$ , and $c=\frac{4x^2-4x+1}{2x^2-5x+2}$

I'm taking your problem to be $\frac{a}{b \times c}$ If so, you can use topher0805's suggestion and rewrite this as $a \times ( \frac{1}{b \times c})$ 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?

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