Solving (x^2+16)(x-1)(x+2)>0 Inequality In Interval Notation

Hey guys! Today, we're diving into a fun little math problem: solving the inequality $(x^2 + 16)(x - 1)(x + 2) > 0$. This might look intimidating at first, but don't worry, we'll break it down step by step. We'll cover everything from identifying critical points to using interval notation, so by the end of this article, you'll be a pro at tackling inequalities like this one. Let's jump right in!

Understanding the Inequality

To solve the inequality (x^2 + 16)(x - 1)(x + 2) > 0, we need to find the values of x that make the expression greater than zero. Inequalities like this are common in algebra and calculus, and mastering them is super important for higher-level math. First things first, let's talk about the different parts of this inequality. We have a product of three factors: $(x^2 + 16)$, $(x - 1)$, and $(x + 2)$. The key to solving this inequality is figuring out when this entire product is positive. Remember, a product is positive if all factors are positive, or if there is an even number of negative factors. The factor $(x^2 + 16)$ is particularly interesting. Notice that $x^2$ is always non-negative for any real number x, so adding 16 makes this factor always positive. This means $(x^2 + 16)$ will not affect the sign of the overall expression, simplifying our problem a bit. Now, we're left with the factors $(x - 1)$ and $(x + 2)$, which are linear factors. These factors can change signs depending on the value of x, and this is where the critical points come into play. Understanding the behavior of these factors will help us determine the intervals where the inequality holds true. We'll find these intervals by identifying the points where the factors are zero or undefined, and then testing values within each interval. This method allows us to see whether the expression is positive or negative in each region, leading us to the solution set. So, let's move on to finding those critical points and see how they help us solve this inequality!

Identifying Critical Points

Critical points are the values of x that make the factors in our inequality equal to zero. These points are crucial because they divide the number line into intervals where the expression's sign remains constant. To find the critical points for our inequality, $(x^2 + 16)(x - 1)(x + 2) > 0$, we focus on the factors that can change signs: $(x - 1)$ and $(x + 2)$. The factor $(x^2 + 16)$ is always positive, as we discussed, so it doesn't give us any critical points. Let's set each of the other factors equal to zero and solve for x: For the factor $(x - 1)$, we have: $x - 1 = 0$. Adding 1 to both sides gives us $x = 1$. This is our first critical point. For the factor $(x + 2)$, we have: $x + 2 = 0$. Subtracting 2 from both sides gives us $x = -2$. This is our second critical point. So, our critical points are $x = 1$ and $x = -2$. These points divide the number line into three intervals: $(-\infty, -2)$, $(-2, 1)$, and $(1, \infty)$. Now, we need to test a value from each interval to see whether the entire expression $(x^2 + 16)(x - 1)(x + 2)$ is positive or negative in that interval. This will help us determine which intervals satisfy the inequality $(x^2 + 16)(x - 1)(x + 2) > 0$. By identifying these critical points, we've taken a significant step toward solving the inequality. Next, we'll use these points to analyze the intervals and find our solution.

Analyzing Intervals

Now that we've identified the critical points $x = -2$ and $x = 1$, we'll analyze the intervals they create to determine where the inequality $(x^2 + 16)(x - 1)(x + 2) > 0$ holds true. These critical points divide the number line into three intervals: $(-\infty, -2)$, $(-2, 1)$, and $(1, \infty)$. To analyze each interval, we'll pick a test value within the interval and plug it into the expression $(x^2 + 16)(x - 1)(x + 2)$. Since $(x^2 + 16)$ is always positive, we only need to consider the signs of $(x - 1)$ and $(x + 2)$.

1. Interval $(-\infty, -2)$:

   Let's choose a test value, say $x = -3$. Plugging this into our factors, we get:

   • $(x - 1) = (-3 - 1) = -4$ (negative)
   • $(x + 2) = (-3 + 2) = -1$ (negative)

   Since both factors are negative, their product is positive. The always-positive factor $(x^2 + 16)$ doesn't change the sign, so the entire expression is positive in this interval. Therefore, the interval $(-\infty, -2)$ satisfies the inequality.

2. Interval $(-2, 1)$:

   Let's choose a test value, say $x = 0$. Plugging this into our factors, we get:

   • $(x - 1) = (0 - 1) = -1$ (negative)
   • $(x + 2) = (0 + 2) = 2$ (positive)

   One factor is negative, and the other is positive, so their product is negative. Again, the positive factor $(x^2 + 16)$ doesn't change the sign, so the entire expression is negative in this interval. Therefore, the interval $(-2, 1)$ does not satisfy the inequality.

3. Interval $(1, \infty)$:

   Let's choose a test value, say $x = 2$. Plugging this into our factors, we get:

   • $(x - 1) = (2 - 1) = 1$ (positive)
   • $(x + 2) = (2 + 2) = 4$ (positive)

   Both factors are positive, so their product is positive. The positive factor $(x^2 + 16)$ keeps the expression positive. Therefore, the interval $(1, \infty)$ satisfies the inequality.

By analyzing these intervals, we've determined where the inequality $(x^2 + 16)(x - 1)(x + 2) > 0$ holds. Now, we'll write our solution in interval notation to clearly represent the set of values that satisfy the inequality.

Expressing the Solution in Interval Notation

Alright, we've done the heavy lifting! We know the intervals where the inequality $(x^2 + 16)(x - 1)(x + 2) > 0$ is true. Now, we just need to express our solution using interval notation. This notation is a concise way to represent sets of numbers, and it's super handy for inequalities. From our analysis, we found that the inequality holds true in the intervals $(-\infty, -2)$ and $(1, \infty)$. Remember, interval notation uses parentheses and brackets to indicate whether the endpoints are included in the solution. Since our inequality is strictly greater than zero ($> 0$), we don't include the endpoints where the expression equals zero. This means we'll use parentheses for both $-2$ and $1$. So, the interval $(-\infty, -2)$ represents all numbers less than $-2$, and the interval $(1, \infty)$ represents all numbers greater than $1$. To combine these intervals into a single solution, we use the union symbol, which looks like this: $\cup$. This symbol means