# How do you simplify (9x^2-16)/ (6x^2-11x+4) div ( 6x^2+11x+4)/(8x^2+10x+3)?

Jul 30, 2015

$E = \frac{9 \cdot \left(x + \frac{3}{4}\right)}{\left(x - \frac{1}{2}\right)}$

#### Explanation:

So, your starting expression looks like this

$E = \frac{9 {x}^{2} - 16}{6 {x}^{2} - 11 x + 4} \cdot \frac{8 {x}^{2} + 10 x + 3}{6 {x}^{2} + 11 x + 4}$

You can use the quadratic formula to factor the denominators of the two fractions, plus the numerator of the second one.

For the general form quadratic $a {x}^{2} + b x + c = 0$, you have

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

You will get

$6 {x}^{2} - 11 x + 4 = 0$

${x}_{1 , 2} = \frac{11 \pm \sqrt{{\left(- 11\right)}^{2} - 4 \cdot 6 \cdot 4}}{2 \cdot 6}$

${x}_{1 , 2} = \frac{11 \pm 5}{12} = \left\{\begin{matrix}{x}_{1} = \frac{11 + 5}{12} = \frac{4}{3} \\ {x}_{2} = \frac{11 - 5}{12} = \frac{1}{2}\end{matrix}\right.$

Notice that $6 {x}^{2} + 11 + 4 = 0$ only differs from the previous quadratic by the sign of the $x$-term, which means that you can write

${x}_{1 , 2} = \frac{- 11 \pm 5}{12} = \left\{\begin{matrix}{x}_{1} = \frac{- 11 - 5}{12} = - \frac{4}{3} \\ {x}_{2} = \frac{- 11 + 5}{12} = - \frac{1}{2}\end{matrix}\right.$

FInally,

$8 {x}^{2} + 10 x + 3 = 0$

${x}_{1 , 2} = \frac{- 10 \pm \sqrt{{10}^{2} - 4 \cdot 8 \cdot 3}}{2 \cdot 8}$

${x}_{1 , 2} = \frac{- 10 \pm 2}{16} = \left\{\begin{matrix}{x}_{1} = \frac{- 10 - 2}{16} = - \frac{3}{4} \\ {x}_{2} = \frac{- 10 + 2}{16} = - \frac{1}{2}\end{matrix}\right.$

Your original expression will now become

$E = \frac{9 {x}^{2} - 16}{\left(x - \frac{1}{2}\right) \cdot \left(x - \frac{4}{3}\right)} \cdot \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x + \frac{1}{2}\right)}}} \cdot \left(x + \frac{3}{4}\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x + \frac{1}{2}\right)}}} \cdot \left(x + \frac{4}{3}\right)}$

$E = \frac{9 {x}^{2} - 16}{\left(x - \frac{1}{2}\right) \cdot \left(x - \frac{4}{3}\right)} \cdot \frac{x + \frac{3}{4}}{x + \frac{4}{3}}$

Now focus on the numerator of the first fraction. Notice that if you use $9$ as a common factor, you can write

$9 \cdot \left({x}^{2} - \frac{16}{9}\right)$

which is equivalent to

$9 \cdot \left({x}^{2} - \frac{16}{9}\right) = 9 \cdot \left[{x}^{2} - {\left(\frac{4}{3}\right)}^{2}\right] = 9 \cdot \left(x - \frac{4}{3}\right) \cdot \left(x + \frac{4}{3}\right)$

Plug this into expression $E$ to get

$E = \frac{9 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - \frac{4}{3}\right)}} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x + \frac{4}{3}\right)}}}}}{\left(x - \frac{1}{2}\right) \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - \frac{4}{3}\right)}}}} \cdot \frac{x + \frac{3}{4}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{x + \frac{4}{3}}}}}$

$E = \textcolor{g r e e n}{\frac{9 \cdot \left(x + \frac{3}{4}\right)}{\left(x - \frac{1}{2}\right)}}$