Question

consider the following polynomial inequality.
$x^2(x + 6)(x - 8) > 0$
step 1 of 2: write the polynomial inequality in the form $p(x) < 0, p(x) \leq 0, p(x) > 0,$ or $p(x) \geq 0$; then find the real zeros of $p(x)$.
answer 4 points
separate multiple answers with a comma.

Explanation:

Step1: Identify the form

The given inequality is \(x^{2}(x + 6)(x - 8)>0\), so \(p(x)=x^{2}(x + 6)(x - 8)\) and we need to find real zeros of \(p(x)\).

Step2: Solve for zeros

Set \(p(x) = 0\), i.e., \(x^{2}(x + 6)(x - 8)=0\).
Using the zero - product property (if \(ab = 0\), then either \(a = 0\) or \(b = 0\)), we have:

• 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 \(x = 8\).

Answer:

\(x=-6,0,8\)