# How do you simplify (4x^2+11x+6)/(x^2-x-6) *( x^2+x-12)/(x^2+8+16)?

Jul 25, 2015

Factor all the quadratics and cancel matching terms to find:

$\frac{4 {x}^{2} + 11 x + 6}{{x}^{2} - x - 6} \cdot \frac{{x}^{2} + x - 12}{{x}^{2} + 8 x + 16} = \frac{4 x + 3}{x + 4}$

with exclusions $x \ne - 2$, $x \ne 3$

#### Explanation:

$\frac{4 {x}^{2} + 11 x + 6}{{x}^{2} - x - 6} \cdot \frac{{x}^{2} + x - 12}{{x}^{2} + 8 x + 16}$

$= \frac{\left(4 {x}^{2} + 11 x + 6\right) \left({x}^{2} + x - 12\right)}{\left({x}^{2} - x - 6\right) \left({x}^{2} + 8 x + 16\right)}$

$= \frac{\cancel{\left(x + 2\right)} \left(4 x + 3\right) \cancel{\left(x + 4\right)} \cancel{\left(x - 3\right)}}{\cancel{\left(x - 3\right)} \cancel{\left(x + 2\right)} \cancel{\left(x + 4\right)} \left(x + 4\right)}$

$= \frac{4 x + 3}{x + 4}$

with exclusions $x \ne - 2$, $x \ne 3$

The values $x = - 2$ and $x = 3$ are excluded because if substituted into the original expression, we get $\frac{0}{0}$ which is undefined, but with the simplified expression, we get a well defined value. These are removable singularities of the original expression.

Therefore the simplified expression is not equivalent to the original expression for those values of $x$.

The value $x = - 4$ is not excluded from the equality of expressions, because it is equally a singularity of the original and simplified expressions, namely a simple pole of order $1$.