1. The problem is to solve the inequality $$\frac{(x-1)(x+3)^2}{(x-4)} \leq 0$$ and determine how many solutions satisfy it. 2. Important points to consider are the zeros of the numerator and denominator: - Numerator zeros: $x=1$ and $x=-3$ (with multiplicity 2) - Denominator zero: $x=4$ (excluded from domain) 3. The critical points divide the number line into intervals: $$(-\infty, -3), (-3, 1), (1, 4), (4, \infty)$$ 4. Analyze the sign of each factor in each interval: - $(x-1)$ changes sign at $x=1$ - $(x+3)^2$ is always non-negative and zero only at $x=-3$ - $(x-4)$ changes sign at $x=4$ 5. Sign of numerator: - For $x < -3$, $(x-1)<0$, $(x+3)^2>0$ so numerator $<0$ - At $x=-3$, numerator $=0$ - For $-3 < x < 1$, $(x-1)<0$, $(x+3)^2>0$ so numerator $<0$ - At $x=1$, numerator $=0$ - For $x > 1$, $(x-1)>0$, $(x+3)^2>0$ so numerator $>0$ 6. Sign of denominator: - For $x < 4$, denominator $<0$ - For $x > 4$, denominator $>0$ 7. Determine the sign of the entire expression in each interval: - $(-\infty, -3)$: numerator $<0$, denominator $<0$ so fraction $>0$ - At $x=-3$: numerator $=0$, fraction $=0$ - $(-3, 1)$: numerator $<0$, denominator $<0$ so fraction $>0$ - At $x=1$: numerator $=0$, fraction $=0$ - $(1, 4)$: numerator $>0$, denominator $<0$ so fraction $<0$ - At $x=4$: denominator $=0$, undefined - $(4, \infty)$: numerator $>0$, denominator $>0$ so fraction $>0$ 8. The inequality $$\frac{(x-1)(x+3)^2}{(x-4)} \leq 0$$ holds where the fraction is less than or equal to zero: - At $x=-3$ and $x=1$ fraction equals zero, so include these points. - On $(1,4)$ fraction is negative, so include this interval. 9. The solution set is: $$[-3, -3] \cup [1,4) = \{-3\} \cup [1,4)$$ 10. Counting the number of solutions: - $x=-3$ is one solution. - $x=1$ is one solution. - The interval $(1,4)$ contains infinitely many solutions. Since the question asks "Adalah sebanyak..." (meaning "How many are there?"), and the options are discrete numbers, the number of solutions is infinite, but considering the nature of the question likely means counting distinct roots or critical points where the expression is zero or changes sign. The zeros of the numerator are at $x=-3$ (multiplicity 2) and $x=1$. The expression is zero at $x=-3$ and $x=1$. The expression is negative on $(1,4)$, so all values in this interval satisfy the inequality. Therefore, the number of solutions is infinite, but if counting distinct points where the expression equals zero or changes sign, there are 3 critical points: $-3$, $1$, and $4$ (excluded). Given the options, the best match is D. 3. Final answer: D. 3