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anonymous

  • 5 years ago

(x^2-2x+4/x^2-5x+6)divided by(x^3+8/x^2-9)

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  1. anonymous
    • 5 years ago
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    \[[(x^2-2x+4)\div(x^2-5x+6)] \div[(x^3+8/x^2-9)]\]

  2. anonymous
    • 5 years ago
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    Multiply the second part by the numerator and simplify.

  3. anonymous
    • 5 years ago
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    I factored down to the following: \[[(x-2)(x-2)/(x-6)(x-1)]*[(x-3)(x+3)/(x+2)(x^2-2x+4)]\]

  4. anonymous
    • 5 years ago
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    But there are no like terms, so do you just combine all of the factors into one problem? I got this as my answer so far: \[(x-2)^2(x-3)(x+3)/(x-1)(x-6)(x+2)(x^2-2x+4)\]

  5. anonymous
    • 5 years ago
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    You flip the second part, then multiply them out. Factoring is unnecessary.

  6. anonymous
    • 5 years ago
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    I did flip the terms, but the bases aren't the same.

  7. anonymous
    • 5 years ago
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    You're multiplying. It doesn't matter if the bases are the same. That only matters with addition and subtraction of fractions.

  8. anonymous
    • 5 years ago
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    So my answer is still technically right, I just have to multiply everything out?

  9. anonymous
    • 5 years ago
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    Yes, but I wouldn't.

  10. anonymous
    • 5 years ago
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    Leave it unfactored, flip it, and multiply through.

  11. anonymous
    • 5 years ago
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    But why include fractions that contain the difference and sums of two squares and cubes if you weren't supposed to factor them?

  12. anonymous
    • 5 years ago
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    Because you're not solving for the roots of x. You'd have to multiply it all out all over again anyway, and make things VERY difficult for yourself. Itd be redundant.

  13. anonymous
    • 5 years ago
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    Multiplying all of that out would take an entire page, and isn't factoring simplifying a problem? My instructions say to simplify completely.

  14. anonymous
    • 5 years ago
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    (didn't include that above, which is my bad)

  15. anonymous
    • 5 years ago
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    You have to factor AFTER simplifying, and AFTER multiplying... Otherwise nothing is simplified, and it's redundant and unnecessary.

  16. anonymous
    • 5 years ago
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    That and you didn't factor correctly, so I'd want to avoid that.

  17. anonymous
    • 5 years ago
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    So you are saying I should take \[(x^2-2x+4)(x^2-9)\] and so on?

  18. anonymous
    • 5 years ago
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    Yes. Then post what you get.

  19. anonymous
    • 5 years ago
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    Oh...

  20. anonymous
    • 5 years ago
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    The numerator should be x^4 -2x^3 + 4x^2 -9x^2 +18x -36 simplify and factor that

  21. anonymous
    • 5 years ago
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    \[x^4-2x^3-5x^2-18x-36\] for above then?

  22. anonymous
    • 5 years ago
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    I had -9x^2+4x^2=-5x^2. Does that check out with you?

  23. anonymous
    • 5 years ago
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    Numerator yes. I'm not sure what you mean by the second post though

  24. anonymous
    • 5 years ago
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    For the numerator, it is fine though. That gives me plenty to work with, thank you again :D

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

  26. anonymous
    • 5 years ago
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    Left Side = Right Side when both are evaluated with x=19

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