Question

$$\frac{(x-8)^{4}\cdot(x+2)}{(x+1)^{3}\cdot(5-x)}\ge0$$

Answer

x<=-2;-2<=x<=8

Solution


Multiply both sides by \({(x+1)}^{3}(5-x)\).
\[{(x-8)}^{4}(x+2)\ge 0\]
Solve for \(x\).
\[x=8,-2\]
From the values of \(x\) above, we have these 3 intervals to test.
\[\begin{aligned}&x\le -2\\&-2\le x\le 8\\&x\ge 8\end{aligned}\]
Pick a test point for each interval.
For the interval \(x\le -2\):
Let's pick \(x=-3\). Then, \(\frac{{(-3-8)}^{4}(-3+2)}{{(-3+1)}^{3}(5+3)}\ge 0\).After simplifying, we get \(228.765625\ge 0\), which is
true
.
Keep this interval.
.
For the interval \(-2\le x\le 8\):
Let's pick \(x=0\). Then, \(\frac{{(0-8)}^{4}(0+2)}{{(0+1)}^{3}(5-0)}\ge 0\).After simplifying, we get \(1638.4\ge 0\), which is
true
.
Keep this interval.
.
For the interval \(x\ge 8\):
Let's pick \(x=9\). Then, \(\frac{{(9-8)}^{4}(9+2)}{{(9+1)}^{3}(5-9)}\ge 0\).After simplifying, we get \(-0.00275\ge 0\), which is
false
.
Drop this interval.
.
Therefore,
\[\begin{aligned}&x\le -2\\&-2\le x\le 8\end{aligned}\]