QuestionConsider the following polynomial inequality. $$x^{2}(x+6)(x-3)>0$$
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 $p(x)$.

Studdy Solution
To find the real zeros of $p(x) = x^2(x+6)(x-3)$, we set $p(x) = 0$ and solve for $x$:
$$x^2(x+6)(x-3) = 0$$
This equation is satisfied when any of the factors is zero:
1. $x^2 = 0$ gives $x = 0$.
2. $x+6 = 0$ gives $x = -6$.
3. $x-3 = 0$ gives $x = 3$.
Thus, the real zeros of $p(x)$ are $x = 0, -6, 3$.
The real zeros of the polynomial are $x = 0, -6, 3$.