# How do you simplify (30x^2+ 2x) /( x^2+x-2 ) *( (x+2)(x-2))/(15x^3-30x^2)?

Oct 16, 2015

(2(15x+1))/(15x(x-1)

#### Explanation:

$\frac{30 {x}^{2} + 2 x}{{x}^{2} + x - 2} \cdot \frac{\left(x + 2\right) \left(x - 2\right)}{15 {x}^{3} - 30 {x}^{2}}$

Your first step is to try and simplify the numerators and denominators as much as possible by factoring them.

$30 {x}^{2} + 2 x = 2 x \left(15 x + 1\right)$

${x}^{2} + x - 2 = {x}^{2} + 2 x - x - 2$

$= x \left(x + 2\right) - \left(x + 2\right)$

$= \left(x + 2\right) \left(x - 1\right)$

and

$15 {x}^{3} - 30 {x}^{2} = 15 {x}^{2} \left(x - 2\right)$

The expression can thus be rewritten as

$\frac{2 x \left(15 x + 1\right)}{\left(x + 2\right) \left(x - 1\right)} \cdot \frac{\left(x + 2\right) \left(x - 2\right)}{15 {x}^{2} \left(x - 2\right)}$

Notice that the expressions that can be found both in the numerator, and in the denominator cancel out to give

$\frac{2 x \left(15 x + 1\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x + 2\right)}}} \left(x - 1\right)} \cdot \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x + 2\right)}}} \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{\left(x - 2\right)}}}}{15 {x}^{2} \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{\left(x - 2\right)}}}}$

You are left with

(2color(purple)(cancel(color(black)(x)))(15x+1))/(x-1) * 1/(15x * color(purple)(cancel(color(black)(x)))) = color(green)( (2(15x+1))/(15x(x-1))