Solution We need to solve the inequality:

\[ \solve{ x^4-9x^2& \gt & 0\\ x^2(x^2-9)& \gt & 0\\ x^2(x-3)(x+3)& \gt &0 } \]

Now that we have the factored polynomial, we use the standard inequality methods. In this case, I will create a table and use the sign changes of the terms to get our answer. The zeros of the polynomial (from the above factors) are -3, 0, and 3.

\[ \begin{{array}}{ |c|c|c|c|c| } \hline &\left(-\infty, -3\right)&\left(-3,0\right)&\left(0,3\right) & \left(3,\infty\right)\\ \hline x^2&\mathbb{+}&\mathbb{+}&\mathbb{+}&\mathbb{+}\\ \hline x+3&\mathbb{-}&\mathbb{+}&\mathbb{+}&\mathbb{+}\\ \hline x-3&\mathbb{-}&\mathbb{-}&\mathbb{-}&\mathbb{+}\\ \hline &\downarrow&\downarrow&\downarrow&\downarrow \\ \hline \text{Result:}&\mathbb{+}&\mathbb{-}&\mathbb{-}&\mathbb{+}\\ \hline &{\color{{green}} \checkmark}&{\color{{red}} \times}&{\color{{red}} \times}&{\color{{green}} \checkmark} \\ \hline \end{{array}} \]

Thus, we can now clearly state with confidence that the polynomial is positive on the interval: \(\left(-\infty, -3\right)\cup\left(3,\infty\right)\).