Solving 3(x-11)(x+3)(x+7) ≤ 0 Polynomial Inequality A Step-by-Step Guide

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Polynomial inequalities can seem daunting, but with a systematic approach, they become manageable. This article aims to provide a comprehensive guide on solving polynomial inequalities, focusing on the specific example of 3(x-11)(x+3)(x+7) ≤ 0. We will delve into the step-by-step process, ensuring you grasp the underlying concepts and can confidently tackle similar problems. Our discussion will cover identifying critical points, testing intervals, and expressing the solution in interval notation. Understanding these techniques is crucial for success in algebra and calculus, as polynomial inequalities frequently appear in various mathematical contexts.

Understanding Polynomial Inequalities

Before we dive into the solution, let's clarify what polynomial inequalities are. A polynomial inequality is an inequality that involves a polynomial expression. Solving it means finding the range(s) of values for the variable (in our case, 'x') that make the inequality true. This is a fundamental concept in algebra and has practical applications in fields like physics, engineering, and economics. For instance, in physics, you might use polynomial inequalities to determine the time intervals during which the height of a projectile remains within a certain range. In economics, they can help model profit margins and cost functions.

Our specific inequality, 3(x-11)(x+3)(x+7) ≤ 0, involves a cubic polynomial. To solve it, we need to find the values of 'x' that make the expression less than or equal to zero. The approach we'll use is a classic method that involves finding critical points (where the polynomial equals zero) and then testing intervals between these points to see where the inequality holds true. This method is applicable to a wide range of polynomial inequalities, making it a valuable tool in your mathematical toolkit. By the end of this guide, you'll not only understand how to solve this particular inequality but also how to apply the same principles to other polynomial inequalities.

Step 1: Finding the Critical Points

The first key step in solving the polynomial inequality 3(x-11)(x+3)(x+7) ≤ 0 is to identify the critical points. Critical points are the values of x that make the polynomial equal to zero. These points are crucial because they divide the number line into intervals where the polynomial's sign remains consistent (either positive or negative). To find the critical points, we set each factor of the polynomial equal to zero and solve for x.

In our case, the polynomial is already factored, which simplifies the process significantly. The factors are: 3, (x - 11), (x + 3), and (x + 7). Setting each factor to zero gives us:

  • 3 = 0 (This is not possible, as 3 is a constant and never equals zero)
  • x - 11 = 0 => x = 11
  • x + 3 = 0 => x = -3
  • x + 7 = 0 => x = -7

Therefore, the critical points are x = 11, x = -3, and x = -7. These three points are the cornerstones of our solution. They are the points where the polynomial can change its sign. By identifying these critical points, we have effectively broken down the number line into four distinct intervals, each of which we will analyze in the next step. The accuracy of these critical points is paramount, as they dictate the boundaries of our intervals. A mistake here can lead to an incorrect solution set.

Step 2: Testing the Intervals

Now that we've identified the critical points (x = -7, x = -3, and x = 11), the next step in solving the polynomial inequality 3(x-11)(x+3)(x+7) ≤ 0 is to test the intervals created by these points. These critical points divide the number line into four distinct intervals: (-∞, -7), (-7, -3), (-3, 11), and (11, ∞). To determine where the polynomial 3(x-11)(x+3)(x+7) is less than or equal to zero, we will select a test value within each interval and evaluate the polynomial at that value. The sign of the result will tell us whether the polynomial is positive or negative in that entire interval.

Let's go through each interval:

  1. Interval (-∞, -7): Choose a test value, say x = -8. Substitute it into the polynomial: 3(-8 - 11)(-8 + 3)(-8 + 7) = 3(-19)(-5)(-1) = -285. Since the result is negative, the polynomial is negative in this interval.
  2. Interval (-7, -3): Choose a test value, say x = -4. Substitute it into the polynomial: 3(-4 - 11)(-4 + 3)(-4 + 7) = 3(-15)(-1)(3) = 135. Since the result is positive, the polynomial is positive in this interval.
  3. Interval (-3, 11): Choose a test value, say x = 0. Substitute it into the polynomial: 3(0 - 11)(0 + 3)(0 + 7) = 3(-11)(3)(7) = -693. Since the result is negative, the polynomial is negative in this interval.
  4. Interval (11, ∞): Choose a test value, say x = 12. Substitute it into the polynomial: 3(12 - 11)(12 + 3)(12 + 7) = 3(1)(15)(19) = 855. Since the result is positive, the polynomial is positive in this interval.

By testing these intervals, we have mapped out the sign behavior of the polynomial across the entire number line. This information is crucial for identifying the intervals where the inequality holds true.

Step 3: Expressing the Solution in Interval Notation

After testing the intervals, we have determined the sign of the polynomial 3(x-11)(x+3)(x+7) in each interval. Now, we need to express the solution to the inequality 3(x-11)(x+3)(x+7) ≤ 0 in interval notation. Remember, we are looking for the intervals where the polynomial is less than or equal to zero. This means we need to include the intervals where the polynomial is negative and also the critical points where the polynomial equals zero.

From our interval testing, we found that the polynomial is negative in the intervals (-∞, -7) and (-3, 11). Additionally, the polynomial is equal to zero at the critical points x = -7, x = -3, and x = 11. Therefore, we include these points in our solution.

In interval notation, we use brackets [ ] to include endpoints and parentheses ( ) to exclude endpoints. Since our inequality includes