# How do you simplify (6x^2-54x+84)/(8x^2-40x+48) div (x^2+x-56)/(x^2+12x+32)?

Jul 30, 2015

$E = \frac{3}{4} \cdot \frac{x + 4}{x - 3}$

#### Explanation:

Start by writing down your starting expression

$E = \frac{6 {x}^{2} - 54 x + 84}{8 {x}^{2} - 40 x + 48} \cdot \frac{{x}^{2} + 12 x + 32}{{x}^{2} + x - 56}$

For the first fraction, you can factor the numerator by $6$ and the denominator by $8$ to get

$E = \frac{6 \cdot \left({x}^{2} - 9 x + 14\right)}{8 \cdot \left({x}^{2} - 5 x + 6\right)} \cdot \frac{{x}^{2} + 12 x + 32}{{x}^{2} + x - 56}$

All these quadratics can be easily factored by using the sum/product technique to get

${x}^{2} - 9 x + 14 = {x}^{2} - 2 x - 7 x + 14 = \left(x - 2\right) \left(x - 7\right)$

${x}^{2} - 5 x + 6 = {x}^{2} - 2 x - 3 x + 6 = \left(x - 2\right) \left(x - 3\right)$

${x}^{2} + 12 x + 32 = {x}^{2} + 4 x + 8 x + 32 = \left(x + 8\right) \left(x + 4\right)$

${x}^{2} + x - 56 = {x}^{2} + 8 x - 7 x + 56 = \left(x + 8\right) \left(x - 7\right)$

The expression will thus be equal to

$E = \frac{6 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 2\right)}}} \cdot \textcolor{g r e e n}{\cancel{\textcolor{b l a c k}{\left(x - 7\right)}}}}{8 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 2\right)}}} \cdot \left(x - 3\right)} \cdot \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{\left(x + 8\right)}}} \cdot \left(x + 4\right)}{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{\left(x + 8\right)}}} \cdot \textcolor{g r e e n}{\cancel{\textcolor{b l a c k}{\left(x - 7\right)}}}}$

E = color(green)(3/4 * (x+4)/(x-3)