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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    Is your problem like this\[x^3-\frac{8}{x^3}-2x^2\]?

  2. anonymous
    • 5 years ago
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    its (x^3-8)/(x^3-2x^2) sorry

  3. anonymous
    • 5 years ago
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    \[\frac{x^3-8}{x^3-2x^2}=\frac{(x-2)(x^2+2x+4)}{x^2(x-2)}=\frac{x^2+2x+4}{x^2}\]\[=\frac{x^2}{x^2}+\frac{2x}{x^2}+\frac{4}{x^2}=1+\frac{2}{x}+\frac{4}{x^2}\]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.

  4. anonymous
    • 5 years ago
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    can you explaine a little more. thats exactly what i want but.. what was your processs

  5. anonymous
    • 5 years ago
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    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?

  6. anonymous
    • 5 years ago
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    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?

  7. anonymous
    • 5 years ago
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    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.

  8. anonymous
    • 5 years ago
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    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).

  9. anonymous
    • 5 years ago
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    oh now i got it thanks a lot! its makes more sense after you use the those number to represent the formula thank you

  10. anonymous
    • 5 years ago
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    np :) become a fan :p

  11. anonymous
    • 5 years ago
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    yes i did

  12. anonymous
    • 5 years ago
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    one more question..how would you devide a function

  13. anonymous
    • 5 years ago
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    Well, it depends on what you're dividing by and what your function is.

  14. anonymous
    • 5 years ago
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    for example (x-3/2x+1)/(2x/2x+1)

  15. anonymous
    • 5 years ago
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    Okay, you have the following,\[\frac{\frac{x-3}{2x+1}}{\frac{2x}{2x+1}}\]

  16. anonymous
    • 5 years ago
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    yup!

  17. anonymous
    • 5 years ago
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    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\[\frac{(x-3)(2x+1)}{(2x)(2x+1)}\]

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

  19. anonymous
    • 5 years ago
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    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.

  20. anonymous
    • 5 years ago
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    oh ok what if denomiator is different

  21. anonymous
    • 5 years ago
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    I'm not sure what you mean.

  22. anonymous
    • 5 years ago
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    just multiply it?

  23. anonymous
    • 5 years ago
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    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 \[\frac{4}{5}\]

  24. anonymous
    • 5 years ago
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    So in that case, you'd just multiply.

  25. anonymous
    • 5 years ago
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    Is that okay, Lammy?

  26. anonymous
    • 5 years ago
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    ok! thanks a lot

  27. anonymous
    • 5 years ago
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    thats help!

  28. anonymous
    • 5 years ago
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    Good. Have fun with it.

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