Question

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

Answer 1

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

Answer 2

We are asked to solve the inequality $\frac{(x-1)(x+2)^2}{(1-x)}<0$. We can rewrite the denominator $1-x$ as $-(x-1)$. So the inequality becomes $\frac{(x-1)(x+2)^2}{-(x-1)}<0$. For $x \neq 1$, we can cancel the term $(x-1)$ from the numerator and denominator. This simplifies the inequality to $\frac{(x+2)^2}{-1} < 0$, which is equivalent to $-(x+2)^2 < 0$. Multiplying by $-1$ and reversing the inequality sign, we get $(x+2)^2 > 0$. This inequality is true for all real numbers $x$ except when $x+2=0$, which means $x=-2$. We also must consider the restriction from the original denominator that $1-x \neq 0$, so $x \neq 1$. Therefore, the solution set includes all real numbers except $x=-2$ and $x=1$. In interval notation, this is $x \in (-\infty, -2) \cup (-2, 1) \cup (1, \infty)$.