Problem 38 - 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. $$ x^{3}+7 x^{2}-x-7<0 $$

Step-by-Step Solution

Answer

The solution to the inequality \(x^{3}+7 x^{2}-x-7<0\) is \((-7,-1) \cup (1,∞)\) in interval notation.

Step 1: Finding the Roots

First, it is necessary to find the roots of the polynomial, as they determine the intervals to be checked. Set the polynomial equal to zero and solve for \(x\). This gives us \(x^{3}+7 x^{2}-x-7=0\). By factoring, or using polynomial division, and then solving using the quadratic equation, we find that \(x=-7, x=1\) and \(x=-1\). These are the roots of the polynomial.

Step 2: Determining Intervals

Now, we need to verify the polynomial's value at the intervals delimited by the roots found in step 1. For our function the intervals are \((-∞,-7)\), \((-7,-1)\), \((-1,1)\) and \((1,∞)\). Choose a test value from each interval and substitute it into the function: if the outcome is negative, the whole interval satisfies the inequality; if the outcome is positive, the inequality is not satisfied on that interval. From this step we found that the function is less than zero on the intervals \((-7,-1)\) and \((1,∞)\)

Step 3: Graphing the Solution

Lastly, we draw these intervals on a real number line - putting marks at \(x=-7\), \(x=-1\), and \(x=1\). We also shade the region where the inequality is true, which is the intervals \((-7,-1)\) and \((1,∞)\). Since the original inequality is a 'less than' inequality without the 'equal to', the solutions \(x=-7\), \(x=-1\), and \(x=1\) are not solutions to the inequality, so the points representing these solutions are open (not filled in) on the graph.