Question: $(x+3)(x-2) \geq 0$
$x^2 + x - 6 \geq 0$
$x+3=0$ ; $x-2=0$
$x_1=-3$ $x_2=2$
$(-\infty,-3]$ $[-3,2]$ $[2,\infty)$
$-3$ $0$ $3$
Interval | Test Point
$(-\infty,-3]$ | $-4$
$[-3,2]$ | $0$
$[2,\infty)$ | $3$
$(x+3)(x-2)\geq0 = (-4+3)(-4-2) = (-1)(-6) = 6 > 0$
true
$= (0+3)(0-2) = 3(-2) = -6 < 0$
false
$= (3+3)(3-2) = 6(1) = 6 > 0$
True
the solution set is $(-\infty,-3] \cup [2,\infty)$
or ${x | x \leq -3}$ or ${x | x \geq 2}$
1. **State the problem:** Solve the inequality $$ (x+3)(x-2) \geq 0 $$ which is equivalent to $$ x^2 + x - 6 \geq 0 $$.
2. **Find the roots:** Set each factor equal to zero:
$$ x+3=0 \Rightarrow x=-3 $$
$$ x-2=0 \Rightarrow x=2 $$
These roots divide the number line into three intervals: $$ (-\infty,-3], [-3,2], [2,\infty) $$.
3. **Test each interval:** Choose a test point in each interval to check the sign of the product.
- For $$ (-\infty,-3] $$, test $$ x=-4 $$:
$$ (x+3)(x-2) = (-4+3)(-4-2) = (-1)(-6) = 6 > 0 $$ so this interval satisfies the inequality.
- For $$ [-3,2] $$, test $$ x=0 $$:
$$ (0+3)(0-2) = 3 \times (-2) = -6 < 0 $$ so this interval does not satisfy the inequality.
- For $$ [2,\infty) $$, test $$ x=3 $$:
$$ (3+3)(3-2) = 6 \times 1 = 6 > 0 $$ so this interval satisfies the inequality.
4. **Include boundary points:** Since the inequality is $$ \geq 0 $$, the roots $$ x=-3 $$ and $$ x=2 $$ where the expression equals zero are included in the solution.
5. **Write the solution set:**
$$ (-\infty,-3] \cup [2,\infty) $$
or in set-builder notation:
$$ \{ x \mid x \leq -3 \} \cup \{ x \mid x \geq 2 \} $$