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

Jan 6, 2017

$\frac{5 \left(b + 5\right)}{b - 1}$

#### Explanation:

First, you can factorize all polynomials:

$5 {b}^{2} - 125 = 5 \left({b}^{2} - 25\right) = 5 \left(b - 5\right) \left(b + 5\right)$

$20 {b}^{2} - 20 = 20 \left({b}^{2} - 1\right) = 20 \left(b - 1\right) \left(b + 1\right)$

$80 b + 80 = 80 \left(b + 1\right)$

$4 b - 20 = 4 \left(b - 5\right)$

Then put them in the given expression:

$\frac{5 \cancel{\left(b - 5\right)} \left(b + 5\right)}{\cancel{20} \left(b - 1\right) \cancel{\left(b + 1\right)}} \cdot \frac{\cancel{80} \cancel{\left(b + 1\right)}}{\cancel{4} \cancel{\left(b - 5\right)}}$

$\frac{5 \left(b + 5\right)}{b - 1}$