anonymous
  • anonymous
how would you solve 4x^2-9/x^2-11x-60 multiplied by x^2-16/2x+3
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Mertsj
  • Mertsj
\[\frac{4x^2-9}{x^2-11x-60} \times \frac{x^2-16}{2x+3}\]
Mertsj
  • Mertsj
\[\frac{(2x-3)(2x+3)}{(x-15)(x+4)} \times \frac{(x-4)(x+4)}{2x+3}\]
ash2326
  • ash2326
\[\frac{4x^2-9}{x^2-11x-60} *\frac{x^2-16}{2x+3}\] 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 \[\frac{(2x-3)(x-4)}{x-15}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Mertsj
  • Mertsj
\[\frac{(2x-3)(x-4)}{x-15}\]
anonymous
  • anonymous
oh ok i forgot to factor the 4x^2-9. Thank you very much!
anonymous
  • anonymous
would the answer for 5x/x^2-7x+10 minus 4/x^2-25 be 5x-4/x-5?
anonymous
  • anonymous
|dw:1328738038677:dw|
anonymous
  • anonymous
complex fraction
ash2326
  • ash2326
\[\frac { \frac{1}{x}-\frac{1}{(x+1)}}{x-1}\] 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 \[\frac 1 {{(x)(x+1)}(x-1)}\]
anonymous
  • anonymous
what happened to the 2?
Mertsj
  • Mertsj
|dw:1328738538884:dw|
Mertsj
  • Mertsj
Final answer: \[\frac{-1}{x(x+1)}\]
anonymous
  • anonymous
oh ok now isee it Thanks again you guys have been a great help
anonymous
  • anonymous
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
anonymous
  • anonymous
|dw:1328739265241:dw|
anonymous
  • anonymous
Rationalizing the denominator

Looking for something else?

Not the answer you are looking for? Search for more explanations.