Verified Answer :

  • Rearrange the inequality to get \(x^3 + 3x^2 - x - 3 \leq 0\).
  • Factor the polynomial by grouping: \((x - 1)(x + 1)(x + 3) \leq 0\).
  • Determine the roots \(x = 1, -1, -3\) and test the intervals created by these points.
  • Identify the solution as the union of intervals where the expression is negative or zero: \(\boxed{(-\infty, -3] \cup [-1, 1]}\).

Explanation

  1. Rearranging the Inequality

We are given the inequality \(x^3 + 3x^2 \leq x + 3\). To solve this, we first bring all terms to one side of the inequality to set it in the form \(P(x) \leq 0\):
\[x^3 + 3x^2 - x - 3 \leq 0\]
This allows us to analyze the behavior of the polynomial \(P(x) = x^3 + 3x^2 - x - 3\) relative to zero.

  1. Factoring the Polynomial

Next, we factor the polynomial \(P(x) = x^3 + 3x^2 - x - 3\) by grouping:
\[x^2(x + 3) - 1(x + 3) \leq 0\]
We can factor out the common term \((x + 3)\):
\[(x^2 - 1)(x + 3) \leq 0\]
Since \(x^2 - 1\) is a difference of squares, we can factor it further as \((x - 1)(x + 1)\). Thus, the fully factored inequality is:
\[(x - 1)(x + 1)(x + 3) \leq 0\]

  1. Finding Roots and Dividing the Number Line

To determine the intervals where the inequality holds, we identify the roots of the polynomial \(P(x) = (x - 1)(x + 1)(x + 3)\). The roots are the values of \(x\) that make the polynomial equal to zero:
\(x - 1 = 0 \implies x = 1\)
\(x + 1 = 0 \implies x = -1\)
\(x + 3 = 0 \implies x = -3\)
These roots divide the real number line into four intervals: \((-\infty, -3]\), \([-3, -1]\), \([-1, 1]\), and \([1, \infty)\).

  1. Testing Intervals

We now test a value in each interval to determine the sign of \(P(x) = (x - 1)(x + 1)(x + 3)\):

  • For the interval \((-\infty, -3]\), pick \(x = -4\): \((-4-1)(-4+1)(-4+3) = (-5)(-3)(-1) = -15 \leq 0\) (True).
  • For the interval \([-3, -1]\), pick \(x = -2\): \((-2-1)(-2+1)(-2+3) = (-3)(-1)(1) = 3 \leq 0\) (False).
  • For the interval \([-1, 1]\), pick \(x = 0\): \((0-1)(0+1)(0+3) = (-1)(1)(3) = -3 \leq 0\) (True).
  • For the interval \([1, \infty)\), pick \(x = 2\): \((2-1)(2+1)(2+3) = (1)(3)(5) = 15 \leq 0\) (False).

Thus, the inequality holds for \(x \in (-\infty, -3] \cup [-1, 1]\).

  1. Conclusion

Based on our analysis, the inequality \(x^3 + 3x^2 \leq x + 3\) is satisfied when \(x\) is in the interval \((-\infty, -3] \cup [-1, 1]\).
Therefore, the solution set is \(\boxed{(-\infty, -3] \cup [-1, 1]}\).

Examples

Polynomial inequalities appear in physics when calculating the possible positions of a particle constrained by energy levels. For instance, if the energy of a particle is given by a cubic function, the regions where the kinetic energy is non-negative are determined by solving an inequality similar to this one. By finding the valid intervals for the position \(x\), scientists can determine where the particle is physically allowed to exist in space.

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