Question

$\frac{(x-1)^2(x-3)(x+3)}{(x+4)^3.x(x-3)}>0$

• A. $\frac{(x-1)^2(x+3)}{(x+4)^3.(x-0)}>0, x\neq 3.$
• B. The solution set is $(-4, -3) \cup (0, 1) \cup (1, 3) \cup (3, \infty)$.
• C. The solution set is $(-4, -3) \cup (0, 3) \cup (3, \infty)$.
• D. The solution set is $(-4, -3) \cup (0, \infty)$.

Answer 1

Answer: (B)

The solution set is $(-4, -3) \cup (0, 1) \cup (1, 3) \cup (3, \infty)$.

Answer 2

The critical points are $x = -4, -3, 0, 1, 3$. The expression simplifies to $\frac{(x-1)^2(x+3)}{x(x+4)^3}$ for $x \neq 3$. We analyze the sign of this expression in the intervals defined by the critical points.

• For $x > 3$, all factors are positive, so the expression is positive.
• For $1 < x < 3$, $(x-1)^2 > 0$, $(x+3) > 0$, $x > 0$, $(x+4)^3 > 0$. The expression is positive.
• For $0 < x < 1$, $(x-1)^2 > 0$, $(x+3) > 0$, $x > 0$, $(x+4)^3 > 0$. The expression is positive.
• For $-3 < x < 0$, $(x-1)^2 > 0$, $(x+3) > 0$, $x < 0$, $(x+4)^3 > 0$. The expression is negative.
• For $-4 < x < -3$, $(x-1)^2 > 0$, $(x+3) < 0$, $x < 0$, $(x+4)^3 > 0$. The expression is positive.
• For $x < -4$, $(x-1)^2 > 0$, $(x+3) < 0$, $x < 0$, $(x+4)^3 < 0$. The expression is negative. The inequality holds for $(-4, -3) \cup (0, 1) \cup (1, 3) \cup (3, \infty)$.