Problem 30 - Algebra and Trigonometry

Question

Solve each polynomial inequality in Exercises \(1-42\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ (x+1)(x+2)(x+3) \geq 0 $$

Step-by-Step Solution

Answer

The solution set is \[ -3, -2 \] and \[ -1, +\infty \].

Step 1: Factor the Polynomial

The given polynomial is \( (x+1)(x+2)(x+3) \geq 0 \). Since it's already factored, this step has nothing else to be done.

Step 2: Determine the Critical Points

Set each of the factored parts equal to 0 and solve for \( x \) to get the critical points. Thus, \( x+1=0 \) gives \( x=-1 \), \( x+2=0 \) gives \( x=-2 \), and \( x+3=0 \) gives \( x=-3 \). So the critical points are \( x=-1, -2, -3 \). These points divide the number line into four intervals: \(-\infty, -3\), \(-3, -2\), \(-2, -1\), \(-1, +\infty\).

Step 3: Test the Intervals

Substitute any number from each interval into the polynomial to check the sign of the polynomial in that interval. If \( x <-3 \), using \( x=-4 \), checking \( (-4+1)(-4+2)(-4+3) \), its negative. If \( -3 < x < -2 \), using \( x = -2.5 \), checking \( (-2.5+1)(-2.5+2)(-2.5+3) \), it's positive. If \( -2 < x <-1 \), using \( x = -1.5 \), checking \( (-1.5+1)(-1.5+2)(-1.5+3) \), it's negative. If \( x > -1 \), using \( x = 0 \) and checking \( (0+1)(0+2)(0+3) \), it's positive.

Step 4: Write the Solution in Interval Notation

The polynomial is greater or equal to 0 for intervals where the test number resulted in a positive sign. These intervals are \[ -3, -2 \] and \[ -1, +\infty \). The square brackets mean that the solutions include the endpoints \( -3, -2 \) and \( -1 \), because the given inequality is greater than or equal to 0.

Step 5: Graph the Solution Set on a Real Number Line

Draw a number line, and shade the intervals \[ -3, -2 \] and \[ -1, +\infty \), and include the points \( -3, -2, -1 \) in the shading since they are part of the solution set.