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anonymous
 5 years ago
reduce each rational expression to lowest terms.
x^38/x^32x^2
anonymous
 5 years ago
reduce each rational expression to lowest terms. x^38/x^32x^2

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is your problem like this\[x^3\frac{8}{x^3}2x^2\]?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its (x^38)/(x^32x^2) sorry

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{x^38}{x^32x^2}=\frac{(x2)(x^2+2x+4)}{x^2(x2)}=\frac{x^2+2x+4}{x^2}\]\[=\frac{x^2}{x^2}+\frac{2x}{x^2}+\frac{4}{x^2}=1+\frac{2}{x}+\frac{4}{x^2}\]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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you explaine a little more. thats exactly what i want but.. what was your processs

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The numerator, x^38, is a difference of two cubes. A difference of two cubes can be factored as\[a^3b^3=(ab)(a^2+ab+b^2)\] 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 \[x^2(x2)\]This means then you have a common factor of (x2) in the numerator and denominator, which cancel to 1. You're left with what I gave you. Is that okay?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0still 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
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, it's a difference of two cubes because x is raised to the power of 3, and 8 is 2^3. So you have\[x^38=x^32^3=(x2)(x^2+2x+2^2)\]using the formula I gave you.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There's a factor of (x2) on the top, and one (x2) on the bottom. They divide out to equal 1 (i.e. they cancel).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh now i got it thanks a lot! its makes more sense after you use the those number to represent the formula thank you

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0np :) become a fan :p

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one more question..how would you devide a function

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, it depends on what you're dividing by and what your function is.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for example (x3/2x+1)/(2x/2x+1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, you have the following,\[\frac{\frac{x3}{2x+1}}{\frac{2x}{2x+1}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This 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 \[a=x3, b=2x+1, c=2x,d=2x+1\]so you can write\[\frac{(x3)(2x+1)}{(2x)(2x+1)}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There's a common factor in the numerator and denominator, 2x+1. This factor will cancel, and you're left with,\[\frac{x3}{2x}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Don'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
 5 years ago
Best ResponseYou've already chosen the best response.0oh ok what if denomiator is different

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm not sure what you mean.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, 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
 5 years ago
Best ResponseYou've already chosen the best response.0So in that case, you'd just multiply.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Good. Have fun with it.
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