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anonymous

  • 4 years ago

how would you solve 4x^2-9/x^2-11x-60 multiplied by x^2-16/2x+3

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  1. Mertsj
    • 4 years ago
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    \[\frac{4x^2-9}{x^2-11x-60} \times \frac{x^2-16}{2x+3}\]

  2. Mertsj
    • 4 years ago
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    \[\frac{(2x-3)(2x+3)}{(x-15)(x+4)} \times \frac{(x-4)(x+4)}{2x+3}\]

  3. ash2326
    • 4 years ago
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    \[\frac{4x^2-9}{x^2-11x-60} *\frac{x^2-16}{2x+3}\] let's factor we know \[a^2-b^2=(a+b)(a-b)\] so we have now \[\frac{(2x-3)(2x+3)}{x^2-11x-60} *\frac{(x-4)(x+4)}{2x+3}\] let's factor x^2-11x-60 let's find factor of -60 , whose sum is -11 -15 and 4 x^2-15x+4x-60 (x-15)(x+4) now let's rewrite thr polynomial fraction given \[\frac{(2x-3)(2x+3)}{(x-15)(x+4)} *\frac{(x-4)(x+4)}{2x+3}\] let's cancel the common terms \[\frac{(2x-3)\cancel{(2x+3)}}{(x-15)\cancel {(x+4)}} *\frac{(x-4)\cancel{(x+4)}}{\cancel{2x+3}}\] we get finally \[\frac{(2x-3)(x-4)}{x-15}\]

  4. Mertsj
    • 4 years ago
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    \[\frac{(2x-3)(x-4)}{x-15}\]

  5. anonymous
    • 4 years ago
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    oh ok i forgot to factor the 4x^2-9. Thank you very much!

  6. anonymous
    • 4 years ago
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    would the answer for 5x/x^2-7x+10 minus 4/x^2-25 be 5x-4/x-5?

  7. anonymous
    • 4 years ago
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    |dw:1328738038677:dw|

  8. anonymous
    • 4 years ago
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    complex fraction

  9. ash2326
    • 4 years ago
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    \[\frac { \frac{1}{x}-\frac{1}{(x+1)}}{x-1}\] LCM of x and x+1 is x(x+1) \[\frac { \frac{x+1}{x(x+1)}-\frac{x}{x(x+1)}}{x-1}\] we get now \[\frac { \frac{1}{x(x+1)}}{x-1}\] we get finally \[\frac 1 {{(x)(x+1)}(x-1)}\]

  10. anonymous
    • 4 years ago
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    what happened to the 2?

  11. Mertsj
    • 4 years ago
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    |dw:1328738538884:dw|

  12. Mertsj
    • 4 years ago
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    Final answer: \[\frac{-1}{x(x+1)}\]

  13. anonymous
    • 4 years ago
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    oh ok now isee it Thanks again you guys have been a great help

  14. anonymous
    • 4 years ago
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    this is all part of this welcome back worksheet in pre cal and i'm not really good with complex fractions. I still have sin,cos,tan on my mind lol

  15. anonymous
    • 4 years ago
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    |dw:1328739265241:dw|

  16. anonymous
    • 4 years ago
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    Rationalizing the denominator

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