If $x^3 + 5x^2 + 8x + 4 = (x + 2)^2(x + 1)$, find the range of values of $x$ for which $\frac{18}{1-x} \ge x^2 + 6x + 14$

1. Solve the inequality from the first part by expressing it as a single rational fraction.

$\frac{18}{1-x} - (x^2 + 6x + 14) \ge 0$
$\frac{18 - (1-x)(x^2 + 6x + 14)}{1-x} \ge 0$
$\frac{18 - (-x^3 - 5x^2 - 8x + 14)}{1-x} \ge 0$
$\frac{x^3 + 5x^2 + 8x + 4}{1-x} \ge 0$
Using the given identity $x^3 + 5x^2 + 8x + 4 = (x + 2)^2(x + 1)$:
$\frac{(x + 2)^2(x + 1)}{1-x} \ge 0$

2. Find the range of values of $x$ for the first part.

The critical values are $x = -2$, $x = -1$, and $x = 1$.
Since $(x + 2)^2 \ge 0$ for all real $x$, the sign of the expression depends on $\frac{x+1}{1-x}$.
For $\frac{x+1}{1-x} \ge 0$, we have $-1 \le x < 1$.
Additionally, the expression equals $0$ when $(x + 2)^2 = 0 \implies x = -2$.
Thus, the solution to the first inequality is $-1 \le x < 1$ or $x = -2$.

3. Substitute $x$ with $|x|$ to find the range of values for the target inequality.

The target inequality is obtained by replacing $x$ with $|x|$ in the first inequality. Therefore, the solution is:
$-1 \le |x| < 1 \quad \text{or} \quad |x| = -2$
Since $|x| \ge 0$ for all real $x$, the equation $|x| = -2$ has no real solutions.
The inequality $-1 \le |x| < 1$ simplifies to $|x| < 1$ (as $-1 \le |x|$ is always true).
$|x| < 1 \implies -1 < x < 1$