How do you simplify #(5b^2-125)/(20b^2-20)*(80b+80)/(4b-20)#?

Question

1466 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-06T07:57:55

#(5(b+5))/(b-1)#

First, you can factorize all polynomials:

#5b^2-125=5(b^2-25)=5(b-5)(b+5)#

#20b^2-20=20(b^2-1)=20(b-1)(b+1)#

#80b+80=80(b+1)#

#4b-20=4(b-5)#

Then put them in the given expression:

#(5cancel((b-5))(b+5))/(cancel20(b-1)cancel((b+1)))*(cancel80cancel((b+1)))/(cancel4cancel((b-5)))#

#(5(b+5))/(b-1)#