Question

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

Answer 1

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

Answer 2

\[\frac{(2x-3)(2x+3)}{(x-15)(x+4)} \times \frac{(x-4)(x+4)}{2x+3}\]

Answer 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
\[\frac{(2x-3)(x-4)}{x-15}\]

Answer 4

\[\frac{(2x-3)(x-4)}{x-15}\]

Answer 5

oh ok i forgot to factor the 4x^2-9. Thank you very much!

Answer 6

would the answer for 5x/x^2-7x+10 minus 4/x^2-25 be 5x-4/x-5?