Solving (x+1)(x-3)^2(x-8)<0: A Step-by-Step Guide

Hey guys! Today, we're diving into a fun little math problem: solving the inequality $(x+1)(x-3)^2(x-8)<0$. Don't worry, it's not as scary as it looks! We'll break it down step-by-step so you can totally nail it. Understanding inequalities is super crucial, especially when you're dealing with functions and their behavior. This particular problem involves a polynomial inequality, which means we need to find the values of x that make the expression less than zero. So, grab your thinking caps, and let’s get started!

1. Understanding Polynomial Inequalities

Before we jump into solving this specific inequality, let's quickly recap what polynomial inequalities are all about. In essence, polynomial inequalities are mathematical statements that compare a polynomial expression to zero. This comparison can be less than (<), greater than (>), less than or equal to (≤), or greater than or equal to (≥). Our goal is to find the range of x-values that satisfy this condition.

The general form of a polynomial inequality looks something like this: $P(x) < 0$, $P(x) > 0$, $P(x) ≤ 0$, or $P(x) ≥ 0$, where $P(x)$ is a polynomial. To solve these, we use critical values – the points where the polynomial equals zero or is undefined – to divide the number line into intervals. Then, we test each interval to see if it satisfies the inequality. This might sound complicated, but trust me, it’s a straightforward process once you get the hang of it. Understanding polynomial inequalities is fundamental in various fields, including calculus and real analysis, making it a core concept in mathematics. These inequalities help us define intervals where functions exhibit specific behaviors, such as increasing or decreasing.

2. Finding the Critical Values

Okay, let's tackle our inequality: $(x+1)(x-3)^2(x-8)<0$. The first step is to find the critical values. These are the values of x that make the expression equal to zero. Think of them as the turning points where the inequality might switch from being true to false, or vice versa. To find these critical values, we set each factor in the inequality to zero:

• $x + 1 = 0$ gives us $x = -1$
• $(x - 3)^2 = 0$ gives us $x = 3$
• $x - 8 = 0$ gives us $x = 8$

So, our critical values are -1, 3, and 8. Notice that $(x-3)$ is squared, which means the root x = 3 has a multiplicity of 2. This will be important later when we analyze the intervals. The critical values act as boundaries on the number line, dividing it into intervals where the polynomial’s sign remains constant. Essentially, these values are the points where the function may change from being positive to negative or vice versa. Identifying critical values is a pivotal step in solving inequalities, as it lays the groundwork for the interval testing that follows. By finding these values, we pinpoint the potential transition points, making it easier to determine the solution intervals.

3. Setting Up the Intervals

Now that we have our critical values, we can set up the intervals on the number line. These values divide the number line into four intervals:

• $(-\infty, -1)$
• $(-1, 3)$
• $(3, 8)$
• $(8, \infty)$

These intervals are crucial because the expression $(x+1)(x-3)^2(x-8)$ will have the same sign throughout each interval. Our next step is to test a value within each interval to see if the inequality holds true. Essentially, we're creating a sign chart to track where the expression is positive or negative. Setting up these intervals correctly is crucial for the subsequent steps in solving the inequality. The number line is effectively partitioned into manageable segments, each bounded by the critical values. Understanding the importance of these intervals allows us to systematically analyze the behavior of the polynomial expression across its entire domain. Each interval represents a potential solution set, which we will explore in the next step.

4. Testing the Intervals

This is where the magic happens! We'll pick a test value from each interval and plug it into our inequality $(x+1)(x-3)^2(x-8)<0$ to see if it holds true. Let's go through each interval:

• Interval $(-\infty, -1)$: Let's pick $x = -2$. Plugging it in, we get $(-2+1)(-2-3)^2(-2-8) = (-1)(25)(-10) = 250$, which is not less than 0. So, this interval is not part of the solution.
• Interval $(-1, 3)$: Let's pick $x = 0$. Plugging it in, we get $(0+1)(0-3)^2(0-8) = (1)(9)(-8) = -72$, which is less than 0. This interval is part of the solution!
• Interval $(3, 8)$: Let's pick $x = 4$. Plugging it in, we get $(4+1)(4-3)^2(4-8) = (5)(1)(-4) = -20$, which is less than 0. This interval is also part of the solution!
• Interval $(8, \infty)$: Let's pick $x = 9$. Plugging it in, we get $(9+1)(9-3)^2(9-8) = (10)(36)(1) = 360$, which is not less than 0. So, this interval is not part of the solution.

Testing the intervals is a critical step in determining which ranges of x-values satisfy the inequality. This process involves substituting a test value from each interval into the original inequality and observing the sign of the result. The sign of the result tells us whether that particular interval is part of the solution set. For instance, if substituting a value makes the inequality true, then the entire interval is included in the solution. This methodical approach ensures that we cover all possible solution sets. The choice of test values within each interval is arbitrary; any value within the interval will yield the same sign for the expression, simplifying the process significantly.

5. Writing the Solution in Interval Notation

Alright, we've done the heavy lifting! Now, let's write the solution in interval notation. We found that the inequality $(x+1)(x-3)^2(x-8)<0$ is true for the intervals $(-1, 3)$ and $(3, 8)$.

However, notice that $x=3$ makes the expression equal to zero, not less than zero. So, we don't include 3 in our solution. Therefore, the solution in interval notation is:

$(-1, 3) \cup (3, 8)$

This means the solution includes all x-values between -1 and 3, and all x-values between 3 and 8, but not -1, 3, or 8 themselves. Writing the solution in interval notation is a concise way to express the set of values that satisfy the inequality. This notation uses parentheses and brackets to indicate whether endpoints are included or excluded. A parenthesis, such as in $(-1, 3)$, indicates that the endpoint is not included, while a bracket would indicate inclusion. When there are multiple disjoint intervals in the solution, we use the union symbol (∪) to combine them, as seen in our final answer. The ability to express solutions in interval notation is a fundamental skill in mathematics, particularly in algebra and calculus, allowing for clear and unambiguous communication of solution sets.

6. A Quick Note on Multiplicity

Remember how we mentioned the multiplicity of the root x = 3? The factor $(x-3)^2$ has a multiplicity of 2, which means the graph of the polynomial touches the x-axis at x = 3 but doesn't cross it. This is why the sign of the expression doesn't change around x = 3. In other words, the interval $(3, 8)$ continues to satisfy the inequality, even though 3 is a critical value. Understanding multiplicity is crucial for accurately interpreting the behavior of polynomial functions near their roots. When a root has an even multiplicity, the function will touch the x-axis at that point but not cross it, meaning the sign of the function does not change across that root. Conversely, a root with an odd multiplicity will result in the function crossing the x-axis, leading to a change in sign. Recognizing the impact of multiplicity allows us to make informed decisions when solving inequalities and analyzing polynomial graphs.

Conclusion

So there you have it! Solving the inequality $(x+1)(x-3)^2(x-8)<0$ might seem tricky at first, but by breaking it down into manageable steps – finding critical values, setting up intervals, testing those intervals, and writing the solution in interval notation – it becomes much easier. Remember, practice makes perfect, so try tackling similar problems to build your skills. You've got this! And that’s a wrap for today, guys. Keep those math muscles flexed, and I'll catch you in the next problem!