solve inequality x^3+3x^2 greater than or equal to x+3



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.

solve inequality x^3+3x^2 greater than or equal to x+3

Learn how to solve the cubic inequality x^3 + 3x^2 ≥ x + 3 using factoring and sign analysis. Follow a detailed step-by-step process that includes factoring the expression, identifying critical points, and testing intervals. Suitable for high school students in Grades 10-12.

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