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Chapter 2: Problem 8

Solve each polynomial inequality and graph the solution set on a real number line. Express each solution set in interval notation. $$x^{2}+x-6>0$$

Short Answer

Expert verified
The solution to \(x^{2}+x-6>0\) is \((- \infty, -3) \cup (2, \infty)\)

Step by step solution

01

Factor the polynomial

First set the inequality equation to equal zero. In doing so, consider it as \(x^{2}+x-6 = 0\). Now factor this equation: \((x-2)(x+3) = 0\). The roots of this equation are \(x = 2\) and \(x = -3\), which form potential boundaries of the intervals in which we check the sign of the inequality.
02

Test each interval

Here we will be determining the intervals by taking the real number line and breaking it up at our calculated roots. This will result in 3 intervals. The intervals are \(-\infty, -3\), \(-3, 2\), and \(2, \infty\). Now we will select a test point inside each interval and substitute into the original inequality. From these calculations, we will be able to determine whether this interval satisfies the inequality. If it does, we may include it in our solution set. For our test points we can select -4, 0, and 3 respectively.
03

Determine the solution set and express it in interval notation

By substituting our test points into the original inequality, we get: \(-4^{2}-4-6>0\) - this is true so \(-\infty, -3\) is part of the solution. For \(0^{2}+0-6>0\), this is false so \(-3, 2\) is not part of the solution. Lastly, \(3^{2}+3-6 > 0\) - true, therefore the interval \(2, \infty\) is part of our solution. Combining these intervals in interval notation gives the solution set as \((- \infty, -3) \cup (2, \infty)\).
04

Graph the solution set on a number line

To graph the solution set on the number line, start by marking the points -3 and 2 on the number line. The solution set \((- \infty, -3)\) corresponds to a line stretching from -3 to the left (negative) infinity, with an opened circle at -3. The solution set \((2, \infty)\) corresponds to a line stretching from 2 to the positive infinity, with an opened circle at 2. As a result, our graph will cover all real numbers less than -3 or greater than 2.

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