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Chapter 2: Problem 3

Solve each polynomial inequality and graph the solution set on a real number line. Express each solution set in interval notation. $$(x-7)(x+3) \leq 0$$

Short Answer

Expert verified
The solution set of the polynomial inequality is \([-3,7]\) union \([7, \infty)\).

Step by step solution

01

Identify the roots

Find the values of \(x\) for which the inequality equals zero. This requires setting each factor of the polynomial to zero. \(x-7=0 \Rightarrow x=7\) \(x+3=0 \Rightarrow x=-3\)
02

Setting up the number line

The number line is divided into 3 intervals by the points \(x=7\) and \(x=-3\). The intervals are: \((- \infty,-3)\), \((-3,7)\), \((7, \infty)\)
03

Test each interval

Select a test point in each interval and substitute it into the inequality. If the inequality is satisfied, then all values in that interval are part of the solution set. Test \(x = -4\) : \((-4-7)\((-4)+3)\) = 7, which is \(> 0\) Test \(x = 0\) : \((0-7)\((0)+3)\) = 21, which is \(> 0\) Test \(x = 8\) : \((8-7)\((8)+3)\) = -5, which is \(< 0\) In the intervals \((- \infty,-3)\) and \((-3,7)\), the inequality is not satisfied, while it is satisfied in the interval \((7, \infty)\).
04

Include Endpoints

The inequality is \(\leq 0\), so it includes the values of \(x\) where the polynomial equals zero. Therefore, the solution includes the points \(x=-3\) and \(x=7\).

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Most popular questions from this chapter

Will help you prepare for the material covered in the next section. Simplify: \(\frac{x+1}{x+3}-2\)

a. If \(y=k x^{2},\) find the value of \(k\) using \(x=2\) and \(y=64\). b. Substitute the value for \(k\) into \(y=k x^{2}\) and write the resulting equation. c. Use the equation from part (b) to find \(y\) when \(x=5\).

Use transformations of \(f(x)=\frac{1}{x}\) or \(f(x)=\frac{1}{x^{2}}\) to graph each rational function. $$h(x)=\frac{1}{(x-3)^{2}}+2$$

What is a rational inequality?

Use inspection to describe each inequality's solution set. Do not solve any of the inequalities. $$(x-2)^{2}>0$$
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