Solving Inequalities: A Step-by-Step Guide

Hey math enthusiasts! Today, we're diving into a classic algebra problem: solving the inequality $\frac{(x - 1)(x-2)(x - 3)}{(x+1)(x+2)(x + 3)} > 1$. Don't worry, it might look a bit intimidating at first, but we'll break it down step by step, making sure you understand every single move. This type of problem tests your understanding of inequalities, rational expressions, and how to handle them correctly. Let's get started and conquer this math challenge together! It is very important to follow each step to make sure that the solution is correct. This is why we break it down into smaller parts.

Step 1: Bringing Everything to One Side and Simplifying

Our first goal is to get all the terms on one side of the inequality. This allows us to compare the expression to zero, which is crucial for determining the solution intervals. We start by subtracting 1 from both sides. This gives us:

$\frac{(x - 1)(x-2)(x - 3)}{(x+1)(x+2)(x + 3)} - 1 > 0$

Next, we need to combine the two terms into a single fraction. To do this, we find a common denominator, which in this case, is $(x+1)(x+2)(x+3)$. So we rewrite the inequality as:

$\frac{(x - 1)(x-2)(x - 3) - (x+1)(x+2)(x + 3)}{(x+1)(x+2)(x + 3)} > 0$

Now we expand the numerator. Expanding $(x - 1)(x-2)(x - 3)$ gives us $x^3 - 6x^2 + 11x - 6$. Similarly, expanding $(x+1)(x+2)(x + 3)$ gives us $x^3 + 6x^2 + 11x + 6$. Subtracting the second expansion from the first, we get:

$\frac{(x^3 - 6x^2 + 11x - 6) - (x^3 + 6x^2 + 11x + 6)}{(x+1)(x+2)(x + 3)} > 0$

Simplifying the numerator, we have $-12x^2 - 12$, which results in:

$\frac{-12x^2 - 12}{(x+1)(x+2)(x + 3)} > 0$

Let's also factor out -12 from the numerator: $\frac{-12(x^2 + 1)}{(x+1)(x+2)(x + 3)} > 0$. Note that the $x^2 + 1$ term is always positive because $x^2$ is always non-negative, and adding 1 makes it strictly positive. This simplification is very important to make sure the next steps are going in the right direction, and is why the problem is broken down.

Step 2: Identifying Critical Points

Critical points are the values of $x$ where the expression either equals zero or is undefined. In this case, the numerator $-12(x^2 + 1)$ can never be zero because $x^2 + 1$ is always positive. However, the denominator $(x+1)(x+2)(x + 3)$ becomes zero when $x = -1, -2,$ or $-3$. These are our critical points.

Also, these critical points divide the number line into intervals. Our critical points are extremely important for determining these intervals, and solving the inequality. These points are not included in the solution because the inequality is strictly greater than zero. This gives us the intervals $(-\,\infty, -3)$, $(-3, -2)$, $(-2, -1)$, and $(-1, \,\infty)$. We will test the sign of the expression within each interval to determine the solution.

Step 3: Creating a Sign Chart

To solve this inequality, we create a sign chart to keep track of the sign of the expression in each interval. The critical points are very useful for making sure the process is right. Because we are dealing with a rational expression, we must also consider where the denominator is zero, as the function will be undefined there.

Our sign chart will have these intervals: $(-\,\infty, -3)$, $(-3, -2)$, $(-2, -1)$, and $(-1, \,\infty)$. We can pick a test value within each interval and substitute it into the simplified expression, $\frac{-12(x^2 + 1)}{(x+1)(x+2)(x + 3)}$. For example, we can use the test values $x = -4, -2.5, -1.5, 0$:

• For $x = -4$: $\frac{-12((-4)^2 + 1)}{(-4+1)(-4+2)(-4 + 3)} = \frac{-12(17)}{(-3)(-2)(-1)} = \frac{-204}{-6} = 34 > 0$ (Positive)
• For $x = -2.5$: $\frac{-12((-2.5)^2 + 1)}{(-2.5+1)(-2.5+2)(-2.5 + 3)} = \frac{-12(7.25)}{(-1.5)(-0.5)(0.5)} = \frac{-87}{0.375} = -232 < 0$ (Negative)
• For $x = -1.5$: $\frac{-12((-1.5)^2 + 1)}{(-1.5+1)(-1.5+2)(-1.5 + 3)} = \frac{-12(3.25)}{(-0.5)(0.5)(1.5)} = \frac{-39}{-0.375} = 104 > 0$ (Positive)
• For $x = 0$: $\frac{-12(0^2 + 1)}{(0+1)(0+2)(0 + 3)} = \frac{-12}{6} = -2 < 0$ (Negative)

Step 4: Determining the Solution

Now, let's analyze the sign chart. We are looking for the intervals where the expression is greater than zero (positive). Based on our tests, the solution intervals are $(-\infty, -3)$ and $(-2, -1)$. This means that the inequality holds true for all $x$ values in these intervals.

Thus, the solution to the inequality $\frac{(x - 1)(x-2)(x - 3)}{(x+1)(x+2)(x + 3)} > 1$ is $x \in (-\,\infty, -3) \cup (-2, -1)$. We exclude the critical points because the inequality is strictly greater than 1, not greater than or equal to 1. Also, remember that the expression is undefined at $x = -1, -2,$ and $-3$ due to the denominator.

Step 5: Verification and Final Thoughts

Verification is an essential step. It's always a good idea to check your solution by testing values within the solution intervals. For example, if we choose $x = -4$ (from $(-\infty, -3)$), the original inequality becomes $\frac{(-4 - 1)(-4-2)(-4 - 3)}{(-4+1)(-4+2)(-4 + 3)} = 34 > 1$, which is true. Likewise, if we choose $x = -1.5$ (from $(-2, -1)$), the original inequality becomes $\frac{(-1.5 - 1)(-1.5-2)(-1.5 - 3)}{(-1.5+1)(-1.5+2)(-1.5 + 3)} = 104 > 1$, which is also true. This helps confirm our solution. It is important to test within the critical points to make sure that the solution is correct.

So, there you have it! We've successfully solved the inequality. Remember, solving inequalities involves several steps, including simplifying the expression, finding critical points, using a sign chart, and testing intervals. This methodical approach ensures you find the correct solution and avoid common pitfalls. Keep practicing, and you'll get better with each problem. Mathematics can be challenging at times. However, you need to keep working through the steps. These concepts build on each other, so it's important to fully understand what the problem is asking. Great job!