How do you simplify [(2x^3+3x^2-2x)/( 3x-15) ÷ (2x^3-x^2)/ (x^2-3x-10)] * (5x^2-10x)/(3x^2+12x+12)?

Jul 30, 2015

Answer:

$E = \frac{5}{18} \cdot \left(x - 2\right)$

Explanation:

Start by writing out your initial expression

$E = \frac{2 {x}^{3} + 3 {x}^{2} - 2 x}{3 x - 15} \cdot \frac{{x}^{2} - 3 x - 10}{2 {x}^{3} - {x}^{2}} \cdot \frac{5 {x}^{2} - 10 x}{3 {x}^{2} + 12 x + 12}$

You can factor the quadratics by using the sum/product technique to get

${x}^{2} - 3 x - 10 = {x}^{2} + 2 x - 5 x - 10 = \left(x - 5\right) \left(x + 2\right)$

and

$3 \left({x}^{2} + 4 x + 4\right) = 3 \left({x}^{2} + 2 x + 2 x + 4\right) = 3 \left(x + 2\right) \left(x + 2\right)$

Your expression can thus be written as

$E = \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{x}}} \left(2 {x}^{2} + 3 x - 2\right)}{3 \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 5\right)}}}} \cdot \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 5\right)}}} \textcolor{p u r p \le}{\cancel{\textcolor{b l a c k}{\left(x + 2\right)}}}}{{x}^{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{2}}}} \left(2 x - 1\right)} \cdot \frac{5 \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{x}}} \left(x - 2\right)}{3 \textcolor{p u r p \le}{\cancel{\textcolor{b l a c k}{\left(x + 2\right)}}} \left(x + 2\right)}$

$E = \frac{2 {x}^{2} + 3 x - 2}{3} \cdot \frac{1}{2 x - 1} \cdot \frac{5 \left(x - 2\right)}{3 \left(x + 2\right)}$

The remaining quadratic can be factored by using the quadratic formula

$\textcolor{b l u e}{{x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}}$

${x}_{1 , 2} = \frac{- 3 \pm \sqrt{{3}^{2} - 4 \cdot 2 \cdot \left(- 2\right)}}{2 \cdot 2}$

${x}_{1 , 2} = \frac{- 3 \pm 5}{4} = \left\{\begin{matrix}{x}_{1} = \frac{- 3 - 5}{4} = - 2 \\ {x}_{2} = \frac{- 3 + 5}{4} = \frac{1}{2}\end{matrix}\right.$

This means that you have

$E = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x + 2\right)}}} \left(x - \frac{1}{2}\right)}{3} \cdot \frac{1}{2 x - 1} \cdot \frac{5 \left(x - 2\right)}{3 \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x + 2\right)}}}}$

E = 5/9 * ((x-2)color(red)(cancel(color(black)((x-1/2)))))/(2color(red)(cancel(color(black)((x-1/2))))) = color(green)(5/18 * (x-2)