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

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

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\[\frac{4x^2-9}{x^2-11x-60} \times \frac{x^2-16}{2x+3}\]
\[\frac{(2x-3)(2x+3)}{(x-15)(x+4)} \times \frac{(x-4)(x+4)}{2x+3}\]
\[\frac{4x^2-9}{x^2-11x-60} *\frac{x^2-16}{2x+3}\] 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 \[\frac{(2x-3)(x-4)}{x-15}\]

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Other answers:

\[\frac{(2x-3)(x-4)}{x-15}\]
oh ok i forgot to factor the 4x^2-9. Thank you very much!
would the answer for 5x/x^2-7x+10 minus 4/x^2-25 be 5x-4/x-5?
|dw:1328738038677:dw|
complex fraction
\[\frac { \frac{1}{x}-\frac{1}{(x+1)}}{x-1}\] 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 \[\frac 1 {{(x)(x+1)}(x-1)}\]
what happened to the 2?
|dw:1328738538884:dw|
Final answer: \[\frac{-1}{x(x+1)}\]
oh ok now isee it Thanks again you guys have been a great help
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
|dw:1328739265241:dw|
Rationalizing the denominator

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