1. ## solve the inequality

i am supposed to solve the inequality

$(x-6)^{2}(x+8)<0$ the zero's are 6 and -8

so the answer would be $(\infty ,-8)$?

2. ## Re: solve the inequality

Originally Posted by slapmaxwell1

i am supposed to solve the inequality

$(x-6)^{2}(x+8)<0$ the zero's are 6 and -8

so the answer would be $(\infty ,-8)$?

3. ## Re: solve the inequality

Originally Posted by slapmaxwell1
i am supposed to solve the inequality
$(x-6)^{2}(x+8)<0$ the zero's are 6 and -8
so the answer would be $(\infty ,-8)$?
Use the LaTeX tags [TEX]...[/TEX]
DO NOT USE $$...$$

4. ## Re: solve the inequality

ok thanks....

5. ## Re: solve the inequality

Originally Posted by slapmaxwell1

i am supposed to solve the inequality

$(x-6)^{2}(x+8)<0$ the zero's are 6 and -8

so the answer would be $(\infty ,-8)$?
First of all, you should write your solution regions from smallest to largest.

Anyway, to solve \displaystyle \begin{align*} (x - 6)^2(x + 8) < 0 \end{align*} you are correct that you should check the roots of the equation \displaystyle \begin{align*} (x - 6)^2(x + 8) = 0 \end{align*}, which are \displaystyle \begin{align*} -8, 6 \end{align*}.

The function will only ever stop being negative or positive at the roots. So we need to check each region that are separated by the roots.

When \displaystyle \begin{align*} x = -9 , (x - 6)^2(x + 8) = -225 < 0 \end{align*}

So \displaystyle \begin{align*} x \in (-\infty, -8) \end{align*} is acceptable.

When \displaystyle \begin{align*} x = 0, (x - 6)^2(x + 8) = 288 > 0 \end{align*}

So \displaystyle \begin{align*} x \in (-8, 6) \end{align*} is not acceptable.

When \displaystyle \begin{align*} x = 7, (x - 6)^2(x + 8) = 15 > 0 \end{align*}

So \displaystyle \begin{align*} x \in (6, \infty) \end{align*} is not acceptable.

Therefore, the solution is \displaystyle \begin{align*} x \in (-\infty, 8) \end{align*}.