### Instruction: Solve
\[\left| \frac{3(x + 1)}{x^2 + 2x + 3} \right| \ge 1.\].### Response: I notice that the fraction inside the absolute value has a quadratic expression in the denominator, and I wonder if I can factor it or find its roots.
I try to factor it by looking for two numbers that multiply to 3 and add to 2, but I don't see any.
I use the quadratic formula to find the roots, and I get \[\frac{-2 \pm \sqrt{4 - 12}}{2} = -1 \pm i\sqrt{2},\] where $i$ is the imaginary unit.
Since the roots are complex, I realize that the denominator is always positive for any real value of $x$, so I don't need to worry about it changing sign.
This means that the absolute value of the fraction depends only on the numerator, and I can rewrite the inequality as \[\left| 3(x + 1) \right| \ge x^2 + 2x + 3.\]
To solve this, I need to consider two cases: either $3(x + 1)$ is positive or negative.
If it is positive, then I can drop the absolute value and solve \[3(x + 1) \ge x^2 + 2x + 3.\]
I subtract $3(x + 1)$ from both sides and get \[0 \ge x^2 - x + 0.\]
This is a quadratic inequality, and I can use the quadratic formula to find the roots of the corresponding equation.
The roots are \[\frac{1 \pm \sqrt{1 - 0}}{2} = \frac{1 \pm 1}{2} = 0, 1.\]
Since the quadratic expression is positive outside the roots and negative between them, I know that the inequality is satisfied when $x$ is between 0 and 1, inclusive.
So one part of the solution is $0 \le x \le 1$.
If $3(x + 1)$ is negative, then I need to change the sign of the inequality and flip the fraction when I drop the absolute value.
I get \[-3(x + 1) \ge x^2 + 2x + 3.\]
I add $3(x + 1)$ to both sides and get \[0 \ge x^2 + 5x + 6.\]
I factor the quadratic expression and get \[0 \ge (x + 2)(x + 3).\]
This means that the expression is negative when $x$ is between -3 and -2, inclusive.
So another part of the solution is $-3 \le x \le -2$.
To find the complete solution, I need to combine these two parts.
Since they are disjoint intervals, I can use the union symbol to join them.
The final solution is \[-3 \le x \le -2 \cup 0 \le x \le 1.\]