Solving (x+2)^2(x+5) ≥ 0: A Step-by-Step Guide

Dec 6, 2025 by Andrew McMorgan 47 views

Hey Plastik Magazine readers! Today, we're diving into the world of inequalities with a fun little problem: solving $(x+2)^2(x+5) \geq 0$ . Inequalities might seem intimidating at first, but with a systematic approach, they become much easier to handle. So, grab your thinking caps, and let's get started!

Understanding the Inequality

Before we jump into solving, let's break down what the inequality $(x+2)^2(x+5) \geq 0$ actually means. We're looking for all the values of x that, when plugged into the expression $(x+2)^2(x+5)$ , give us a result that is either greater than or equal to zero. In other words, we want the expression to be non-negative.

The key here is to understand the factors involved: $(x+2)^2$ and $(x+5)$ . Notice that $(x+2)^2$ is a squared term. What does that tell us? Well, any real number squared is always non-negative (i.e., greater than or equal to zero). This is a crucial observation that will simplify our work.

Let's think about $(x+5)$ . This factor will be negative when x is less than -5, zero when x is equal to -5, and positive when x is greater than -5. The interplay between these two factors, $(x+2)^2$ and $(x+5)$ , will determine the sign of the entire expression.

Finding Critical Points

Critical points are the values of x that make the expression $(x+2)^2(x+5)$ equal to zero. These points are important because they divide the number line into intervals where the expression is either positive or negative. To find the critical points, we set each factor equal to zero:

$(x+2)^2 = 0$ implies $x+2 = 0$ , so $x = -2$ .
$(x+5) = 0$ implies $x = -5$ .

Thus, our critical points are x = -2 and x = -5. These are the values where the expression changes its sign or equals zero. Now, let's use these critical points to create intervals on the number line.

Creating Intervals and Testing Values

Our critical points, -5 and -2, divide the number line into three intervals: $(-\infty, -5)$ , $(-5, -2)$ , and $(-2, \infty)$ . We'll test a value from each interval to determine the sign of the expression $(x+2)^2(x+5)$ in that interval.

Interval $(-\infty, -5)$ : Let's pick $x = -6$ . Then, $(x+2)^2(x+5) = (-6+2)^2(-6+5) = (-4)^2(-1) = 16(-1) = -16$ . Since -16 is less than 0, the expression is negative in this interval.
Interval $(-5, -2)$ : Let's pick $x = -3$ . Then, $(x+2)^2(x+5) = (-3+2)^2(-3+5) = (-1)^2(2) = 1(2) = 2$ . Since 2 is greater than 0, the expression is positive in this interval.
Interval $(-2, \infty)$ : Let's pick $x = 0$ . Then, $(x+2)^2(x+5) = (0+2)^2(0+5) = (2)^2(5) = 4(5) = 20$ . Since 20 is greater than 0, the expression is positive in this interval.

Determining the Solution

Remember, we want to find where $(x+2)^2(x+5) \geq 0$ . This means we're looking for the intervals where the expression is either positive or equal to zero. From our testing, we found:

The expression is negative on $(-\infty, -5)$ .
The expression is positive on $(-5, -2)$ .
The expression is positive on $(-2, \infty)$ .

Since we want the expression to be greater than or equal to zero, we also need to include the critical points in our solution. Therefore:

We include -5 because the expression equals zero at $x = -5$ .
We include -2 because the expression equals zero at $x = -2$ .

So, our solution consists of the interval $(-5, -2)$ , the interval $(-2, \infty)$ , and the points -5 and -2. We can write this as a union of intervals:

$[-5, -2] \cup [-2, \infty)$

Notice that $[-5, -2] \cup [-2, \infty)$ is the same as $[-5, \infty)$ . We include -5 because it makes the factor $(x+5)$ zero, and we include -2 because it makes the factor $(x+2)^2$ zero. Both satisfy the greater than or equal to zero condition.

Final Answer

Thus, the solution to the inequality $(x+2)^2(x+5) \geq 0$ is:

$x \in [-5, \infty)$

Visualizing the Solution

It's always a good idea to visualize the solution on a number line. Imagine a number line stretching from negative infinity to positive infinity. Mark the critical points -5 and -2. The solution includes everything from -5 to positive infinity, including -5. This is because at x = -5, the expression equals zero, which satisfies the inequality. x = -2 is included for the same reason: it makes the expression equal zero, fulfilling the condition that the expression must be greater than or equal to zero.

The interval is closed (using a square bracket) at -5, indicating that -5 is part of the solution. The number line extends indefinitely to the right, showing that all values greater than -5 also satisfy the inequality.

Common Mistakes to Avoid

When solving inequalities, there are a few common mistakes that students often make. Here's what to watch out for:

Forgetting to include critical points: Remember to include critical points in your solution if the inequality includes "equal to" ( $\geq$ or $\leq$ ). In our case, we included -5 and -2 because the inequality was $(x+2)^2(x+5) \geq 0$ .
Incorrectly testing intervals: Be careful when testing values in each interval. Make sure you plug the value into the original expression and evaluate it correctly.
Assuming squared terms are always positive: While squared terms are always non-negative, they can be zero. Don't forget to consider when they equal zero, as these points might be part of the solution.
Not understanding interval notation: Familiarize yourself with interval notation. Square brackets [ ] mean the endpoint is included, while parentheses ( ) mean the endpoint is not included.

Additional Tips for Solving Inequalities

Here are a few extra tips that might come in handy when solving inequalities:

Simplify the expression first: If possible, simplify the expression before finding critical points. This might involve factoring or expanding terms.
Consider the behavior of each factor: Analyze each factor separately to understand how it affects the sign of the expression.
Use a sign chart: A sign chart can be helpful for organizing your work and visualizing the sign of the expression in each interval.
Check your answer: After finding the solution, check it by plugging in a few values from the solution set into the original inequality.

Conclusion

Alright guys, solving the inequality $(x+2)^2(x+5) \geq 0$ is a great example of how to break down a problem into smaller, manageable steps. Remember to find the critical points, create intervals, test values, and consider the "equal to" part of the inequality. With practice, you'll become a pro at solving inequalities! Keep rocking those math problems, and we'll catch you in the next one here at Plastik Magazine! Stay curious, stay stylish, and keep those brains buzzing!