Question 77106
Factor the numerator {{{25x^2-16}}}

*[invoke quadratic_factoring 25, 0, -16]

Now factor out an x from the denominator

{{{x(15x^2+7x-4)}}}

Factor the quadratic in the parenthesis

*[invoke quadratic_factoring 15, 7, -4]

So the expression goes from 

{{{(25x^2-16)/(15X^3+7x^2-4x)}}}

to

{{{((5x+4)(5x-4))/(x(3x-1)(5x+4))}}}



{{{(cross(5x+4)(5x-4))/(x(3x-1)cross(5x+4))}}}

Which reduces to 
{{{(5x-4)/(x(3x-1))}}}