To solve the inequality $x^3 + 3x^2 \geq x + 3$, follow these steps:

Step 1: Move all terms to one side of the inequality.

$x^3 + 3x^2 - x - 3 \geq 0$ This simplifies the inequality to: $f(x) = x^3 + 3x^2 - x - 3 \geq 0$

Step 2: Factor the expression.

To factor $f(x) = x^3 + 3x^2 - x - 3$, let's try factoring by grouping: $f(x) = (x^3 + 3x^2) - (x + 3)$ Factor each group: $f(x) = x^2(x + 3) - 1(x + 3)$ Now factor out $(x + 3)$: $f(x) = (x + 3)(x^2 - 1)$

Step 3: Further factor the quadratic term.

Notice that $x^2 - 1$ is a difference of squares: $x^2 - 1 = (x - 1)(x + 1)$ So the full factorization of the expression is: $f(x) = (x + 3)(x - 1)(x + 1)$

Step 4: Solve for the critical points.

To find when the expression equals zero, solve $(x + 3)(x - 1)(x + 1) = 0$: $x + 3 = 0 \quad \Rightarrow \quad x = -3$ $x - 1 = 0 \quad \Rightarrow \quad x = 1$ $x + 1 = 0 \quad \Rightarrow \quad x = -1$ The critical points are $x = -3$, $x = -1$, and $x = 1$.

Step 5: Determine the sign of the expression in each interval.

We now test the sign of $(x + 3)(x - 1)(x + 1)$ in the intervals divided by the critical points $x = -3, -1, 1$.

• For $x < -3$, say $x = -4$: $(-4 + 3)(-4 - 1)(-4 + 1) = (-1)(-5)(-3) = -15 \quad \text{(negative)}$

• For $-3 < x < -1$, say $x = -2$: $(-2 + 3)(-2 - 1)(-2 + 1) = (1)(-3)(-1) = 3 \quad \text{(positive)}$

• For $-1 < x < 1$, say $x = 0$: $(0 + 3)(0 - 1)(0 + 1) = (3)(-1)(1) = -3 \quad \text{(negative)}$

• For $x > 1$, say $x = 2$: $(2 + 3)(2 - 1)(2 + 1) = (5)(1)(3) = 15 \quad \text{(positive)}$

Step 6: Conclusion

We want $(x + 3)(x - 1)(x + 1) \geq 0$, so we are interested in the intervals where the expression is positive or zero. From the sign analysis:

• The expression is positive in the intervals $(-3, -1)$ and $(1, \infty)$.
• The expression is zero at $x = -3$, $x = -1$, and $x = 1$.

Thus, the solution to the inequality is: $x \in [-3, -1] \cup [1, \infty)$

Final Answer:

$x \in [-3, -1] \cup [1, \infty)$

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Would you like any further details or have any questions?

Here are some related questions:

1. How can I solve inequalities with higher powers of $x$?
2. What is the method for factoring cubic polynomials?
3. How do I solve inequalities that involve absolute values?
4. Can inequalities have multiple disconnected solution intervals?
5. How can I graph the solution to a polynomial inequality?

Tip: When solving polynomial inequalities, it's helpful to first solve the equality and then test intervals between the roots to determine where the expression is positive or negative.