To solve $x^2 - x - 6 > 0$ you need every $x$ that makes the expression positive. The zeros are the only places the sign can change, so they split the number line into pieces. This connects to Compound Inequalities: the answer is often an "or."
Key idea
Move everything to one side, find the zeros, and mark them on a number line. The sign is the same throughout each piece, so test one number per piece (or picture the parabola). Keep the pieces with the sign you want; include the zeros only for $\ge$ or $\le$.
Why it works
A polynomial graph is an unbroken curve, so it can't jump from positive to negative without passing through 0. Between two neighboring zeros the curve stays on one side of the $x$-axis. For a parabola that opens up, it is below the axis between its zeros and above outside them.
Worked examples
Worked example: Outside the zeros
Solve $x^2 - x - 6 > 0$.
- Factor: $(x + 2)(x - 3) > 0$. Zeros: $-2$ and $3$.
- Test $x = 0$: $(2)(-3) = -6$, negative. Test $x = 4$: $(6)(1) = 6$, positive. Test $x = -3$: $(-1)(-6) = 6$, positive.
- Keep the positive pieces. The zeros make it 0, which is not $> 0$.
Answer: $x < -2$ or $x > 3$, that is, $(-\infty, -2) \cup (3, \infty)$
Worked example: The trap with squares
Solve $x^2 \le 9$.
- Rewrite: $x^2 - 9 \le 0$, so $(x + 3)(x - 3) \le 0$.
- Between the zeros the product is negative; at the zeros it is 0. Both are allowed.
Answer: $-3 \le x \le 3$
Worked example: A parabola that opens down
Solve $-x^2 + 4x + 5 \ge 0$.
- Factor: $-(x^2 - 4x - 5) = -(x + 1)(x - 5)$.
- This parabola opens down, so it is above the axis between its zeros.
Answer: $-1 \le x \le 5$
Common mistake
Solving $x^2 > 4$ as $x > \pm 2$. That would say $x > -2$, which includes $x = 0$, and $0^2 > 4$ is false. The two zeros point in opposite directions: $x < -2$ or $x > 2$. Always test a number from each piece in the original inequality.