1. **State the problem:** We need to find how many integers satisfy the inequality $$\frac{(x+2)(x+3)}{x-2} \geq 0$$ and are less than 5. 2. **Identify critical points:** The expression is undefined at $x=2$ (denominator zero). The numerator zeros are at $x=-2$ and $x=-3$. 3. **Intervals to test:** The critical points divide the real line into intervals: - $(-\infty, -3)$ - $(-3, -2)$ - $(-2, 2)$ - $(2, \infty)$ 4. **Sign analysis:** - For $x < -3$, pick $x=-4$: numerator $(x+2)(x+3) = (-4+2)(-4+3) = (-2)(-1) = 2 > 0$, denominator $-4-2 = -6 < 0$, so fraction $>0/ <0 = <0$ (negative). - For $-3 < x < -2$, pick $x=-2.5$: numerator $(-2.5+2)(-2.5+3) = (-0.5)(0.5) = -0.25 < 0$, denominator $-2.5-2 = -4.5 < 0$, fraction $<0 / <0 = >0$ (positive). - For $-2 < x < 2$, pick $x=0$: numerator $(0+2)(0+3) = 2*3=6 > 0$, denominator $0-2 = -2 < 0$, fraction $>0 / <0 = <0$ (negative). - For $x > 2$, pick $x=3$: numerator $(3+2)(3+3) = 5*6=30 > 0$, denominator $3-2=1 > 0$, fraction $>0 / >0 = >0$ (positive). 5. **Check zeros:** - At $x=-3$, numerator zero, fraction zero (allowed since inequality is \geq 0). - At $x=-2$, numerator zero, fraction zero (allowed). - At $x=2$, denominator zero, undefined, so $x=2$ is excluded. 6. **Solution intervals:** - $[-3, -2]$ (fraction $\\geq 0$) - $(2, \infty)$ 7. **Find integers less than 5 in solution:** - From $[-3, -2]$: integers $-3, -2$ - From $(2, 5)$: integers $3, 4$ Total integers: $-3, -2, 3, 4$ which is 4 integers. **Final answer:** D. 4