Solve the inequality x^2 - 3x - 10 <= 0.

Step-by-step solution

1. 1

   Factor the left-hand side. Two numbers with product $-10$ and sum $-3$ are $-5$ and $2$:

   $x^2-3x-10=(x-5)(x+2)\le 0$

2. 2

   Find the roots. The product is zero at $x=5$ and $x=-2$. Because the inequality is non-strict, these two values are part of the solution.

3. 3

   Decide which region makes the product negative. The leading coefficient is $+1$, so the parabola opens upward: it is below the axis only between its roots. Equivalently, for $-2<x<5$ the factor $(x+2)$ is positive while $(x-5)$ is negative, giving a negative product.

4. 4

   State the closed interval.

   $-2\le x\le 5,\qquad\text{i.e. }[-2,\,5]$

5. 5

   Test all three regions plus the endpoints. At $x=0$: $-10\le 0$ $\checkmark$. At $x=6$: $36-18-10=8\not\le 0$ $\checkmark$. At $x=-3$: $9+9-10=8\not\le 0$ $\checkmark$. At $x=5$ and $x=-2$ the expression is exactly $0$, which satisfies $\le 0$ $\checkmark$.