How do you multiply #(3x^2-1)/(4x(x-5)) div ((3x^2)(x-1))/(x-5)#?

1 Answer
May 23, 2015

First, remember that when you're dividing two fractions, you invert the other and multiply them instead, like this:

#(3x^2-1)/(4x(x-5))*(x-5)/((3x^2)(x-1))#

We can already cancel some things that are both dividing and multiplying the equation and, thus, having a null effect in it.

#(3x^2-1)/(4xcancel(x-5))*cancel(x-5)/((3x^2)(x-1))#

#(3x^2-1)/(12x^4-12x^3)#