Question

Multiplying rational expressions:
x+5/4x-16 * 2x^2-32/x^2-25

Answer 1

we have
\[\frac{x+5}{4x+16} * \frac{2x^2-32}{x^2-25}\]
now we can take out 4 from 4x +16 and 2 from 2x^2-32
so we get
\[\frac{x+5}{4(x+4)} * \frac{2(x^2-16)}{x^2-25}\]
now using a^-b^2=(a+b)(a-b)
\[\frac{x+5}{4(x+4)} * \frac{2(x+4)(x-4)}{(x-5)(x+5)}\]
now we can cancel x+5 and x+4 from numerator and denominator,
so we have
\[\frac{\cancel{x+5}}{4{\cancel{(x+4)}}} * \frac{2\cancel{(x+4)}(x-4)}{(x-5)\cancel{(x+5)}}\]
so we have now
\[\frac{2(x-4)}{4(x-5)}\]
we get finally
\[\frac{x-4}{2(x-5)}\]

Answer 2

why would we take oout 4 and 2?

Answer 3

I will tell you

if i said .. i give 2 apple for 4 persons
then of course
each person have \[\frac{1}{2}\] apple
its the simplest form while you said it \[\frac{2}{4}, \frac{1}{2}\] both are true ...
The origin of Your Qustion is simplification .. so ash2326 do that in best way .