I know tO factor but I'm not sure here to start
Solve for x:
(16-x^2)/(x^2 +x-6)>or equal to 0
5x^(2)e^x + 2xe^x > 0
Please explain how to solve them, thanks.
$\displaystyle \displaystyle \frac{16 - x^2}{x^2 + x - 6} \geq 0$
First of all, it should be obvious that $\displaystyle \displaystyle x \neq -3$ and $\displaystyle \displaystyle x \neq 2$. (Why?)
You will need to consider two cases, the first where $\displaystyle \displaystyle x^2 + x - 6 < 0$, and the second where $\displaystyle \displaystyle x^2 + x - 6 > 0$.
Case 1: First note that since the denominator is negative, but the entire fraction is nonnegative, that means the numerator must not be positive. So $\displaystyle \displaystyle 16 - x^2 \leq 0 \implies x^2 - 16 \geq 0$.
$\displaystyle \displaystyle \begin{align*}\frac{16 - x^2}{x^2 + x - 6} &\geq 0 \\ 16 - x^2 &\leq 0 \textrm{ (since the denominator is negative, the inequality sign changes)} \\ x^2 - 16 &\geq 0 \\ x^2 &\geq 16 \\ \sqrt{x^2} &\geq \sqrt{16} \\ |x| &\geq 4 \\ x \leq -4 \textrm{ or }x \geq 4 \end{align*}$
Now have a try of Case 2.
Ah, thank you to both of you for your help.
But wouldn't case two be incorrect? Seeing as the numerator must be negative to counter the negative denominator to make the result more than zero. If there were a positive numerator wouldn't that mean the equation as a whole I negative an thus less than zero when it's supposed to be greaterthan zero?
For the second, remember that $\displaystyle \displaystyle e^x > 0$ for all $\displaystyle \displaystyle x$, so
$\displaystyle \displaystyle \begin{align*} 5x^2e^x + 2x\,e^x &> 0 \\ e^x\left(5x^2 + 2x\right) &> 0 \\ 5x^2 + 2x &> 0 \\ x^2 + \frac{2}{5}x &> 0 \\ x^2 + \frac{2}{5}x + \left(\frac{1}{5}\right)^2 &> \left(\frac{1}{5}\right)^2 \\ \left(x + \frac{1}{5}\right)^2 &> \frac{1}{25} \\ \left|x + \frac{1}{5}\right| &> \frac{1}{5} \\ x + \frac{1}{5} < -\frac{1}{5} \textrm{ or } x + \frac{1}{5} &> \frac{1}{5} \\ x < -\frac{2}{5} \textrm{ or } x &> 0\end{align*}$