Question

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

Answer 1

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

Answer 2

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

Answer 3

\[\frac{x^3-8}{x^3-2x^2}=\frac{(x-2)(x^2+2x+4)}{x^2(x-2)}=\frac{x^2+2x+4}{x^2}\]\[=\frac{x^2}{x^2}+\frac{2x}{x^2}+\frac{4}{x^2}=1+\frac{2}{x}+\frac{4}{x^2}\]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.

Answer 4

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

Answer 5

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?

Answer 6

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?

Answer 7

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.

Answer 8

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).

Answer 9

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

Answer 10

np :) become a fan :p

Answer 11

yes i did

Answer 12

one more question..how would you devide a function

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

yup!

Answer 17

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\[\frac{(x-3)(2x+1)}{(2x)(2x+1)}\]

Answer 18

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

Answer 19

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.

Answer 20

oh ok what if denomiator is different

Answer 21

I'm not sure what you mean.

Answer 22

just multiply it?

Answer 23

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 \[\frac{4}{5}\]

Answer 24

So in that case, you'd just multiply.

Answer 25

Is that okay, Lammy?

Answer 26

ok! thanks a lot

Answer 27

thats help!

Answer 28

Good. Have fun with it.