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Chapter 3: Problem 11

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. $$ 3 x^{2}+10 x-8 \leq 0 $$

Short Answer

Expert verified
The solution to the inequality \(3x^2 + 10x - 8 \leq 0\) is \(x \leq 2/3\) and \(x \geq -8/3\). The intervals in which the inequality holds true are \((-\infty, -8/3] \cup [0.7, \infty)\'.

Step by step solution

01

Setting the inequality to zero

Start by setting the inequality to zero to find the critical points: \(3x^2 + 10x - 8 = 0\)
02

Factoring the polynomial

Then, factor the polynomial to solve for x. The factored polynomial is \(3(x - 0.7)(x + 8/3) = 0\)
03

Solving for x

Setting each factor equal to zero gives the solutions: \(x = 0.7\) and \(x = -8/3]\)
04

Testing intervals

Next, test the intervals cut by the roots on the real number line using a number from each of the intervals in the inequality. Negative infinity to -8/3, -8/3 to 0.7, and 0.7 to infinity. If the inequality is true, that interval is part of the solution set. For this polynomial, the solution set in interval notation is: \((-\infty, -8/3] \cup [0.7, \infty)\)
05

Graphing solution on a number line

Finally, draw the solutions on a number line. Show open dots at x = -8/3 and x = 0.7, and a closed dot at x = 0.7 because the inequality is less than OR EQUAL TO 0, and the solution includes x = 0.7.

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