reduce each rational expression to lowest terms.
x^3-8/x^3-2x^2

- anonymous

reduce each rational expression to lowest terms.
x^3-8/x^3-2x^2

- schrodinger

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- anonymous

Is your problem like this\[x^3-\frac{8}{x^3}-2x^2\]?

- anonymous

its (x^3-8)/(x^3-2x^2) sorry

- anonymous

\[\frac{x^3-8}{x^3-2x^2}=\frac{(x-2)(x^2+2x+4)}{x^2(x-2)}=\frac{x^2+2x+4}{x^2}\]\[=\frac{x^2}{x^2}+\frac{2x}{x^2}+\frac{4}{x^2}=1+\frac{2}{x}+\frac{4}{x^2}\]assuming this is what you mean by 'expression to lowest terms'. In fact, you could probably just leave it like (x^2+2x+4)/x^2.

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## More answers

- anonymous

can you explaine a little more. thats exactly what i want but.. what was your processs

- anonymous

The numerator, x^3-8, is a difference of two cubes. A difference of two cubes can be factored as\[a^3-b^3=(a-b)(a^2+ab+b^2)\]
The denominator has a common factor of x^2, and I saw that by taking that factor out, I would have the numerator factored as \[x^2(x-2)\]This means then you have a common factor of (x-2) in the numerator and denominator, which cancel to 1. You're left with what I gave you. Is that okay?

- anonymous

still trying to understand.. =( i got the denominator part.. how did the numerator come to be? (x−2)(x2+2x+4 is it factor too?

- anonymous

Yeah, it's a difference of two cubes because x is raised to the power of 3, and 8 is 2^3. So you have\[x^3-8=x^3-2^3=(x-2)(x^2+2x+2^2)\]using the formula I gave you.

- anonymous

There's a factor of (x-2) on the top, and one (x-2) on the bottom. They divide out to equal 1 (i.e. they cancel).

- anonymous

oh now i got it thanks a lot! its makes more sense after you use the those number to represent the formula thank you

- anonymous

np :) become a fan :p

- anonymous

yes i did

- anonymous

one more question..how would you devide a function

- anonymous

Well, it depends on what you're dividing by and what your function is.

- anonymous

for example (x-3/2x+1)/(2x/2x+1)

- anonymous

Okay, you have the following,\[\frac{\frac{x-3}{2x+1}}{\frac{2x}{2x+1}}\]

- anonymous

yup!

- anonymous

This is the same thing as when you have fractions that you're diving by. When you have two fractions a/b and c/d, and you divide a/b by c/d, you should recall,\[\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}\times \frac{d}{c}=\frac{ad}{bc}\]I won't go into the algebra for why this is the case, since it distracts a little from the point of what you have to do.
Here you have \[a=x-3, b=2x+1, c=2x,d=2x+1\]so you can write\[\frac{(x-3)(2x+1)}{(2x)(2x+1)}\]

- anonymous

There's a common factor in the numerator and denominator, 2x+1. This factor will cancel, and you're left with,\[\frac{x-3}{2x}\]

- anonymous

Don't be put off by the x's and the bits and pieces that are attached to it. They're just numbers in the end.

- anonymous

oh ok what if denomiator is different

- anonymous

I'm not sure what you mean.

- anonymous

just multiply it?

- anonymous

Oh, I see...yes. You can't cancel anything out if it's not the same, just like you can't cancel anything from a fraction like \[\frac{4}{5}\]

- anonymous

So in that case, you'd just multiply.

- anonymous

Is that okay, Lammy?

- anonymous

ok! thanks a lot

- anonymous

thats help!

- anonymous

Good. Have fun with it.

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