# How do you simplify (15-14x-8x^2)/ (4x^2 +4x-15) div ( 4x^2+13x-12)/(3x^2+16x+5)?

Jul 19, 2015

This appears to be intended to be an exercise in factoring:

$15 - 14 x - 8 {x}^{2} = \left(5 + 2 x\right) \left(3 - 4 x\right)$

$4 {x}^{2} + 4 x - 15 = \left(2 x - 3\right) \left(2 x + 5\right)$

$4 {x}^{2} + 13 x - 12 = \left(4 x - 3\right) \left(x + 4\right)$

$3 {x}^{2} + 16 x + 5 = \left(3 x + 1\right) \left(x + 5\right)$

Returning to the original expression:

$\frac{15 - 14 x - 8 {x}^{2}}{4 {x}^{2} + 4 x - 15} \div \frac{4 {x}^{2} + 13 x - 12}{3 {x}^{2} + 16 x + 5}$

$=$$\textcolor{w h i t e}{\text{XXXX}}$(15-14x-8x^2)/(4x^2+4x-15) * (3x^2+16x+5) /(4x^2+13x -12)

$=$$\textcolor{w h i t e}{\text{XXXX}}$$\frac{\left(5 + 2 x\right) \left(3 - 4 x\right)}{\left(2 x - 3\right) \left(2 x + 5\right)} \cdot \frac{\left(3 x + 1\right) \left(x + 5\right)}{\left(4 x - 3\right) \left(x + 4\right)}$

$=$$\textcolor{w h i t e}{\text{XXXX}}$$\frac{\cancel{\left(5 + 2 x\right)} \left(3 - 4 x\right)}{\left(2 x - 3\right) \cancel{\left(2 x + 5\right)}} \cdot \frac{\left(3 x + 1\right) \left(x + 5\right)}{\left(4 x - 3\right) \left(x + 4\right)}$

$=$$\textcolor{w h i t e}{\text{XXXX}}$$\frac{{\cancel{\left(3 - 4 x\right)}}^{- 1}}{\left(2 x - 3\right)} \cdot \frac{\left(3 x + 1\right) \left(x + 5\right)}{\cancel{\left(4 x - 3\right)} \left(x + 4\right)}$

$=$$\textcolor{w h i t e}{\text{XXXX}}$$- \frac{\left(3 x + 1\right) \left(x + 5\right)}{\left(2 x - 3\right) \left(x + 4\right)}$

or
$=$$\textcolor{w h i t e}{\text{XXXX}}$$\frac{3 {x}^{2} + 16 x + 5}{2 {x}^{2} + 5 x - 12}$