Answer
$$[-3,-2]∪[-1+∞)$$
Work Step by Step
Let $(x + 1)(x + 2)(x + 3) \geq 0$ be a function $f$.
Find the $x$-intercepts by solving $(x + 1)(x + 2)(x + 3) \geq 0=0$
$x+1=0$ or $x+2=0$ or $x+3=0$
$x = -1$ or $x =-2$ or $x=-3$
These $x$-intercepts serve as boundary points that separate the number line into intervals.
The boundary points of this equation therefore divide the number line into three intervals:
$(-∞, -3)(-3,-2),(-2,-1),(-1, +∞)$
Choose one test value within each interval and evaluate $f$ at that number.
$(-∞, -3)$
Test Value = $-4$
$f=(x + 1)(x + 2)(x + 3)$
$f=(-4+ 1)(-4 + 2)(-4+ 3)$
$f=-6$
Conclusion: $f (x) < 0$ for all $x$ in $(-∞, -3)$.
$(-3, -2)$
Test Value = $-\frac{5}{2}$
$f=(x + 1)(x + 2)(x + 3)$
$f=(-\frac{5}{2} + 1)(-\frac{5}{2} + 2)(-\frac{5}{2} + 3)$
$f=\frac{3}{8}$
Conclusion: $f (x) > 0$ for all $x$ in $(-3,-2)$.
$(-2, -1)$
Test Value = $-\frac{3}{2}$
$f=(x + 1)(x + 2)(x + 3)$
$f=(-\frac{3}{2} + 1)(-\frac{3}{2} + 2)(-\frac{3}{2} + 3)$
$f=-\frac{3}{8}$
Conclusion: $f (x) < 0$ for all $x$ in $(-2,-1)$.
$(-1,+∞)$
Test Value = $0$
$f=(x + 1)(x + 2)(x + 3)$
$f=(0 + 1)(0 + 2)(0 + 3)$
$f=6$
Conclusion: $f (x) > 0$ for all $x$ in $(-1,+∞)$.
Write the solution set, selecting the interval or intervals that satisfy the given inequality. We are interested in solving $f (x) \geq 0$, where $f (x) = (x + 1)(x + 2)(x + 3)$. Based on the solution above, $f(x)\gt0$ for all $x$ in $(-3,-2)$ or $(-1,+∞)$. However, because the inequality involves $\geq$ (greater than or equal to), we must also include the solutions of $(x + 1)(x + 2)(x + 3) = 0$, namely $-1$, $-2$ and $-3$, in the solution set.
Thus, the solution set of the given inequality is:
$$[-3,-2]∪[-1+∞)$$
