# How do you simplify (x^2-x-12)/(8x^2)div(x^3+3x^2)/(8x^3-2x^2)div(4x-1)/(x+2)?

Aug 3, 2016

$\frac{{x}^{2} - x - 12}{8 {x}^{2}} \div \frac{{x}^{3} + 3 {x}^{2}}{8 {x}^{3} - 2 {x}^{2}} \div \frac{4 x - 1}{x + 2} = \frac{\left(x - 4\right) \left(x + 2\right)}{4 {x}^{2}}$

#### Explanation:

$\frac{{x}^{2} - x - 12}{8 {x}^{2}} \div \frac{{x}^{3} + 3 {x}^{2}}{8 {x}^{3} - 2 {x}^{2}} \div \frac{4 x - 1}{x + 2}$

= $\frac{{x}^{2} - 4 x + 3 x - 12}{8 {x}^{2}} \div \frac{{x}^{2} \left(x + 3\right)}{2 {x}^{2} \left(4 x - 1\right)} \div \frac{4 x - 1}{x + 2}$

= $\frac{x \left(x - 4\right) x + 3 \left(x - 4\right)}{8 {x}^{2}} \div \frac{{x}^{2} \left(x + 3\right)}{2 {x}^{2} \left(4 x - 1\right)} \div \frac{4 x - 1}{x + 2}$

= $\frac{\left(x + 3\right) \left(x - 4\right)}{8 {x}^{2}} \times \frac{2 {x}^{2} \left(4 x - 1\right)}{{x}^{2} \left(x + 3\right)} \times \frac{x + 2}{4 x - 1}$

= (cancel((x+3))(x-4))/(4cancel(8x^2))xx(cancel((2x^2))cancel((4x-1)))/(x^2cancel((x+3)))xx(x+2)/(cancel((4x-1))

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