Question

(w) $\frac{(x-1)(x+2)^2}{(1-x)}<0$

Answer 1

$x \in (-\infty, -2) \cup (-2, 1) \cup (1, \infty)$

Answer 2

The given inequality is $\frac{(x-1)(x+2)^2}{(1-x)}<0$.

Rewrite the denominator: $(1-x) = -(x-1)$. The inequality becomes: $\frac{(x-1)(x+2)^2}{-(x-1)}<0$.

For $x \neq 1$, we can simplify $\frac{(x-1)}{-(x-1)} = -1$. The inequality simplifies to: $-1 \cdot (x+2)^2 < 0$. This is equivalent to: $-(x+2)^2 < 0$.

Multiply by $-1$ and reverse the inequality sign: $(x+2)^2 > 0$.

This inequality is true for all real numbers $x$ except when $(x+2)^2 = 0$, which occurs at $x = -2$. So, the solution is $x \neq -2$.

Considering the original denominator $(1-x)$, we also have the restriction $x \neq 1$.

Combining these conditions, the solution set is all real numbers except $-2$ and $1$. Therefore, $x \in (-\infty, -2) \cup (-2, 1) \cup (1, \infty)$.