# How do you simplify (x^2-x-6)/(4x^3)*(x+1)/(x^2+5x+5)?

Jul 18, 2015

Try factoring and find:

$\frac{{x}^{2} - x - 6}{4 {x}^{3}} \cdot \frac{x + 1}{{x}^{2} + 5 x + 5}$

$= \frac{\left(x - 3\right) \left(x + 2\right) \left(x + 1\right)}{4 {x}^{3} \left(x + \frac{5 + \sqrt{5}}{2}\right) \left(x + \frac{5 - \sqrt{5}}{2}\right)}$ (factoring)

$= \frac{{x}^{3} - 7 x - 6}{4 {x}^{5} + 20 {x}^{4} + 20 {x}^{3}}$ (multiplying)

#### Explanation:

Going in one direction, multiply up to get:

$\frac{{x}^{2} - x - 6}{4 {x}^{3}} \cdot \frac{x + 1}{{x}^{2} + 5 x + 5}$

$= \frac{\left({x}^{2} - x - 6\right) \left(x + 1\right)}{4 {x}^{3} \left({x}^{2} + 5 x + 5\right)}$

$= \frac{{x}^{3} - 7 x - 6}{4 {x}^{5} + 20 {x}^{4} + 20 {x}^{3}}$

Going in the other direction, factor to get:

$\frac{{x}^{2} - x - 6}{4 {x}^{3}} \cdot \frac{x + 1}{{x}^{2} + 5 x + 5}$

$\frac{\left(x - 3\right) \left(x + 2\right)}{4 {x}^{3}} \cdot \frac{x + 1}{{x}^{2} + 5 x + 5}$

$= \frac{\left(x - 3\right) \left(x + 2\right) \left(x + 1\right)}{4 {x}^{3} \left(x + \frac{5 + \sqrt{5}}{2}\right) \left(x + \frac{5 - \sqrt{5}}{2}\right)}$

No common factors to cancel, so this cannot be simplified.