# How do you resolve the following inequality: ((x^2-4)(3x-6))/(x-7)>0?

Apr 19, 2016

$x \in \left(- \infty , - 2\right) \cup \left(7 , \infty\right)$

#### Explanation:

If we factor the numerator of the left hand side completely, we obtain

$\frac{3 \left(x + 2\right) {\left(x - 2\right)}^{2}}{x - 7} > 0$

We wish to identify the intervals on which the left hand side is positive. As the sign can only change at points where the left hand side evaluates to $0$ (a factor of the number is $0$) or is undefined (a factor of the denominator is $0$), we will check the intervals with endpoints at those values.

We can also simplify the task slightly by noting that ${\left(x - 2\right)}^{2} > 0$ for all $x \ne 2$, and so that factor will never change the sign of the expression. Proceeding:

On $\left(- \infty , - 2\right)$ we have $x + 2 < 0$ and $x - 7 < 0$, meaning we have a $\frac{-}{-} = +$ case, and so $\frac{3 \left(x + 2\right) {\left(x - 2\right)}^{2}}{x - 7} > 0$.

On $\left(- 2 , 7\right)$ we have $x + 2 > 0$ and $x - 7 < 0$, meaning we have a $\frac{+}{-} = -$ case, and so $\frac{3 \left(x + 2\right) {\left(x - 2\right)}^{2}}{x - 7} < 0$

(note that the expression is $0$ at $x = 2$, and so $x = 2$ would not be in the solution set in any case)

On $\left(7 , \infty\right)$ we have $x + 2 > 0$ and $x - 7 > 0$, meaning we have a $\frac{+}{+} = +$ case, and so $\frac{3 \left(x + 2\right) {\left(x - 2\right)}^{2}}{x - 7} > 0$.

As we only want the $+$ cases, we get our final solution set as $x \in \left(- \infty , - 2\right) \cup \left(7 , \infty\right)$.