Question

consider the following polynomial inequality.
$x^{2}(x + 6)(x - 8) > 0$
step 2 of 2: test each interval to find the solution of the polynomial inequality.
express your answer in interval notation.
answer 4 points
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Explanation:

Step1: Find Critical Points

To solve the polynomial inequality \(x^{2}(x + 6)(x - 8)>0\), we first find the critical points by setting each factor equal to zero:

• For \(x^{2}=0\), we get \(x = 0\) (with multiplicity 2).
• For \(x + 6=0\), we get \(x=-6\).
• For \(x - 8=0\), we get \(x = 8\).

These critical points divide the number line into the intervals: \((-\infty,-6)\), \((-6,0)\), \((0,8)\), and \((8,\infty)\).

Step2: Test Intervals

We test a value from each interval in the inequality \(x^{2}(x + 6)(x - 8)>0\):

• Interval \((-\infty,-6)\): Let's choose \(x=-7\). Then \(x^{2}=(-7)^{2}=49>0\), \(x + 6=-7 + 6=-1<0\), \(x - 8=-7-8=-15<0\). The product is \(49\times(-1)\times(-15)=735>0\). So this interval is part of the solution.
• Interval \((-6,0)\): Let's choose \(x=-1\). Then \(x^{2}=(-1)^{2}=1>0\), \(x + 6=-1 + 6=5>0\), \(x - 8=-1-8=-9<0\). The product is \(1\times5\times(-9)=-45<0\). So this interval is not part of the solution.
• Interval \((0,8)\): Let's choose \(x = 1\). Then \(x^{2}=1^{2}=1>0\), \(x + 6=1 + 6=7>0\), \(x - 8=1-8=-7<0\). The product is \(1\times7\times(-7)=-49<0\). So this interval is not part of the solution.
• Interval \((8,\infty)\): Let's choose \(x=9\). Then \(x^{2}=9^{2}=81>0\), \(x + 6=9 + 6=15>0\), \(x - 8=9-8=1>0\). The product is \(81\times15\times1 = 1215>0\). So this interval is part of the solution.

We also note that \(x = 0\) makes the left - hand side equal to 0 (since \(x^{2}=0\)), so it is not included in the solution. \(x=-6\) and \(x = 8\) also make the left - hand side equal to 0, so they are not included.

Answer:

\((-\infty,-6)\cup(8,\infty)\)