Solving Polynomial Inequality -u³ ≥ 17u² + 70u Interval And Graphical Solutions

In this article, we will delve into the process of solving the polynomial inequality $-u^3 \geq 17u^2 + 70u$. Polynomial inequalities, a fundamental concept in algebra, involve finding the range of values for a variable that satisfy a given inequality involving polynomials. These inequalities are pervasive in various mathematical and real-world applications, making it crucial to understand the techniques for solving them. Our exploration will encompass algebraic manipulations, factorization, and graphical representation to arrive at the solution, expressing it in both interval notation and graphical form. This comprehensive approach will enhance your understanding of polynomial inequalities and equip you with the skills to tackle similar problems effectively. Specifically, we'll start by rearranging the inequality to set one side to zero, which is a standard first step in solving polynomial inequalities. This rearrangement allows us to identify the critical points, which are the values of u where the polynomial equals zero. These critical points divide the number line into intervals, and we'll test a value within each interval to determine whether the inequality holds true. By systematically analyzing these intervals, we can pinpoint the solution set, which consists of the intervals where the inequality is satisfied. The solution set will then be expressed using interval notation, a concise way of representing a range of values. Finally, we'll visually represent the solution on a number line, providing a graphical interpretation that complements the algebraic solution.

1. Rearranging the Inequality

The first step in solving the polynomial inequality is to rearrange it so that one side is zero. This allows us to work with a standard form and identify the critical points more easily.

We are given the inequality:

$$-u^3 \geq 17u^2 + 70u$$

To rearrange, we add $u^3$ to both sides and subtract $17u^2$ and $70u$ from both sides:

$$0 \geq u^3 + 17u^2 + 70u$$

This can also be written as:

$$u^3 + 17u^2 + 70u \leq 0$$

Rearranging the inequality in this way sets the stage for factoring the polynomial expression. By ensuring that one side of the inequality is zero, we can focus on finding the roots of the polynomial, which are the values of u that make the polynomial equal to zero. These roots serve as critical points that divide the number line into intervals, within which the polynomial's sign remains constant. The process of rearranging the inequality is not just a mathematical manipulation; it's a strategic move that simplifies the problem and makes it more amenable to solution. By bringing all terms to one side, we create a scenario where the polynomial's behavior (whether it's positive, negative, or zero) can be easily analyzed within each interval defined by the roots. This step is crucial because it transforms the inequality problem into a root-finding problem, which is a more familiar and manageable task in algebra. Moreover, rearranging the inequality helps in visualizing the problem graphically. The roots of the polynomial correspond to the points where the graph of the polynomial intersects the x-axis, providing a visual representation of the solution set. Therefore, this initial step is not merely a technicality but a fundamental aspect of solving polynomial inequalities, laying the groundwork for the subsequent steps of factorization and interval analysis.

2. Factoring the Polynomial

Next, we factor the polynomial expression to find its roots. Factoring simplifies the polynomial and allows us to identify the values of u that make the expression equal to zero.

We have:

$$u^3 + 17u^2 + 70u \leq 0$$

We can factor out a common factor of u:

$$u(u^2 + 17u + 70) \leq 0$$

Now, we factor the quadratic expression $u^2 + 17u + 70$. We look for two numbers that multiply to 70 and add to 17. These numbers are 7 and 10. So, we can factor the quadratic as:

$$u(u + 7)(u + 10) \leq 0$$

Factoring the polynomial is a crucial step in solving polynomial inequalities because it transforms a complex expression into a product of simpler factors. This factorization reveals the roots of the polynomial, which are the values of the variable that make the polynomial equal to zero. In this case, factoring out the common factor of u immediately simplifies the expression, making it easier to identify the remaining factors. The subsequent factorization of the quadratic expression $u^2 + 17u + 70$ further breaks down the polynomial into linear factors, each of which corresponds to a root of the polynomial. The process of finding two numbers that multiply to 70 and add to 17 is a classic technique for factoring quadratic expressions. By identifying these numbers, 7 and 10, we can rewrite the quadratic as a product of two binomials, $(u + 7)$ and $(u + 10)$. This complete factorization, $u(u + 7)(u + 10)$, is essential for determining the critical points of the inequality. These critical points are the values of u that make any of the factors equal to zero, and they serve as boundaries that divide the number line into intervals. Within each interval, the sign of the polynomial remains constant, allowing us to determine whether the inequality is satisfied in that interval. Therefore, factoring is not just a mathematical exercise; it's a strategic step that unlocks the structure of the polynomial and paves the way for solving the inequality by identifying the critical points.

3. Identifying Critical Points

The critical points are the values of u that make the polynomial equal to zero. These points divide the number line into intervals.

From the factored form, $u(u + 7)(u + 10) \leq 0$, we set each factor equal to zero to find the critical points:

• $u = 0$
• $u + 7 = 0 \implies u = -7$
• $u + 10 = 0 \implies u = -10$

Thus, the critical points are $u = -10$, $u = -7$, and $u = 0$.

Identifying the critical points is a pivotal step in solving polynomial inequalities because these points serve as the boundaries that divide the number line into distinct intervals. Within each of these intervals, the polynomial's sign remains consistent, meaning it is either strictly positive or strictly negative. The critical points themselves are the values of the variable that make the polynomial equal to zero, and they correspond to the roots of the polynomial equation. In this case, the factored form of the polynomial, $u(u + 7)(u + 10)$, allows us to easily identify the critical points by setting each factor equal to zero. This process yields the critical points $u = -10$, $u = -7$, and $u = 0$, which are the values of u that make the polynomial expression equal to zero. These critical points are not just isolated values; they are the points where the graph of the polynomial intersects the x-axis. Graphically, they represent the x-intercepts of the polynomial function. The critical points are essential for creating a sign chart or a test interval diagram, which is a visual tool used to determine the sign of the polynomial in each interval. By testing a value within each interval, we can determine whether the polynomial is positive or negative in that interval, and this information is crucial for identifying the solution set of the inequality. Therefore, identifying the critical points is not just a procedural step; it's a fundamental aspect of understanding the behavior of the polynomial and determining the intervals where the inequality is satisfied.

4. Creating a Sign Chart

A sign chart helps determine the sign of the polynomial in each interval created by the critical points. We test a value from each interval in the factored inequality $u(u + 7)(u + 10) \leq 0$.

The critical points are $-10$, $-7$, and $0$. These points divide the number line into four intervals:

• $(-\infty, -10)$
• $(-10, -7)$
• $(-7, 0)$
• $(0, \infty)$

We choose test values within each interval:

• For $(-\infty, -10)$, let's choose $u = -11$:

  $$(-11)(-11 + 7)(-11 + 10) = (-11)(-4)(-1) = -44$$

  The result is negative.
• For $(-10, -7)$, let's choose $u = -8$:

  $$(-8)(-8 + 7)(-8 + 10) = (-8)(-1)(2) = 16$$

  The result is positive.
• For $(-7, 0)$, let's choose $u = -1$:

  $$(-1)(-1 + 7)(-1 + 10) = (-1)(6)(9) = -54$$

  The result is negative.
• For $(0, \infty)$, let's choose $u = 1$:

  $$(1)(1 + 7)(1 + 10) = (1)(8)(11) = 88$$

  The result is positive.

Creating a sign chart is a strategic method for visualizing the sign of the polynomial expression within each interval defined by the critical points. This visual representation is essential for determining where the polynomial satisfies the given inequality. The sign chart is constructed by first identifying the critical points, which, as we established earlier, are the roots of the polynomial equation. These critical points divide the number line into distinct intervals, and within each interval, the sign of the polynomial remains constant. To determine the sign of the polynomial in each interval, we select a test value within that interval and substitute it into the factored form of the polynomial. The sign of the resulting expression indicates the sign of the polynomial throughout the entire interval. In this case, the critical points $-10$, $-7$, and $0$ divide the number line into four intervals: $(-\infty, -10)$, $(-10, -7)$, $(-7, 0)$, and $(0, \infty)$. By choosing test values such as $-11$, $-8$, $-1$, and $1$, we can evaluate the sign of the polynomial $u(u + 7)(u + 10)$ in each interval. For example, when $u = -11$, the polynomial evaluates to a negative value, indicating that the polynomial is negative throughout the interval $(-\infty, -10)$. Similarly, we can determine the sign of the polynomial in the other intervals. The sign chart provides a clear and concise summary of the polynomial's sign in each interval, making it straightforward to identify the intervals where the inequality is satisfied. This step is crucial for pinpointing the solution set of the inequality, which consists of the intervals where the polynomial meets the specified condition (in this case, being less than or equal to zero).

5. Determining the Solution

We are looking for where $u(u + 7)(u + 10) \leq 0$. From the sign chart, the inequality is satisfied when the polynomial is negative or zero.

The polynomial is negative in the intervals $(-\infty, -10)$ and $(-7, 0)$. It is equal to zero at the critical points $u = -10$, $u = -7$, and $u = 0$. Thus, the solution includes these points.

In interval notation, the solution is:

$$(-\infty, -10] \cup [-7, 0]$$

Determining the solution set is the culmination of the previous steps, where we synthesize the information gathered from factoring the polynomial, identifying critical points, and creating a sign chart. The sign chart provides a clear visual representation of the intervals where the polynomial expression is positive, negative, or zero. To solve the inequality, we focus on the intervals where the polynomial satisfies the given condition, which in this case is $u(u + 7)(u + 10) \leq 0$. This means we are looking for the intervals where the polynomial is either negative or equal to zero. From the sign chart, we can readily identify the intervals where the polynomial is negative. These are the intervals where the test values resulted in a negative evaluation of the polynomial expression. Additionally, we must include the critical points in the solution set because these are the values of u that make the polynomial equal to zero, which satisfies the "or equal to" part of the inequality. In this specific example, the polynomial is negative in the intervals $(-\infty, -10)$ and $(-7, 0)$. The critical points $u = -10$, $u = -7$, and $u = 0$ are also part of the solution because they make the polynomial equal to zero. To express the solution set in interval notation, we use brackets to include the critical points and parentheses to exclude the endpoints of the intervals. The union symbol $\cup$ is used to combine the disjoint intervals into a single solution set. Therefore, the solution in interval notation is $(-\infty, -10] \cup [-7, 0]$, which represents the range of values of u that satisfy the original polynomial inequality.

6. Graphical Form

The solution in graphical form is represented on a number line. We mark the critical points $-10$, $-7$, and $0$ on the number line. Since the inequality is non-strict ($ \leq $), we use closed circles (or brackets) to indicate that these points are included in the solution. The intervals where the polynomial is negative are shaded.

The graphical representation would show a shaded region from negative infinity up to and including $-10$, a shaded region from $-7$ up to and including $0$, and open (unshaded) regions elsewhere.

The graphical representation of the solution set provides a visual complement to the algebraic solution, enhancing understanding and offering a different perspective on the range of values that satisfy the polynomial inequality. In this graphical representation, we use a number line as the foundation, with the critical points marked as significant landmarks. The critical points, which are the roots of the polynomial equation, divide the number line into intervals, and the solution set consists of the intervals where the polynomial is either negative or zero. To visually represent the solution, we use closed circles (or brackets) on the number line at the critical points to indicate that these points are included in the solution set. This is because the inequality is non-strict, meaning it includes the "equal to" case, so the values of u that make the polynomial equal to zero are part of the solution. The intervals where the polynomial is negative are shaded on the number line, visually highlighting the range of values that satisfy the inequality. In contrast, the intervals where the polynomial is positive are left unshaded, indicating that these values are not part of the solution set. This graphical representation allows for a quick and intuitive understanding of the solution set. It visually demonstrates the continuous range of values that satisfy the inequality, as well as the specific critical points that are included. The shaded regions on the number line provide a clear picture of the solution, making it easier to grasp the overall concept and communicate the solution effectively. The combination of the algebraic solution in interval notation and the graphical representation on a number line offers a comprehensive understanding of the solution to the polynomial inequality.

Conclusion

The solution to the polynomial inequality $-u^3 \geq 17u^2 + 70u$ is given by the interval notation $(-\infty, -10] \cup [-7, 0]$. The graphical form includes closed circles at $-10$, $-7$, and $0$ with the number line shaded between $(-\infty, -10]$ and $[-7, 0]$.

This comprehensive guide has walked you through each step involved in solving the polynomial inequality, from rearranging and factoring to creating a sign chart and determining the solution in both interval notation and graphical form. By understanding these steps, you can confidently tackle similar polynomial inequality problems. This method of solving polynomial inequalities is not just a mathematical exercise; it's a powerful tool with applications in various fields, including physics, engineering, and economics. Polynomial inequalities are used to model real-world situations where constraints or conditions must be met, such as determining the range of values for a variable that satisfies a particular physical law or economic model. The ability to solve these inequalities accurately and efficiently is therefore a valuable skill for anyone working in these fields. The combination of algebraic manipulation, factorization, and graphical representation provides a comprehensive approach to solving polynomial inequalities, allowing for a deeper understanding of the solution set and its implications. The interval notation offers a concise and precise way of expressing the solution, while the graphical representation provides a visual aid that enhances comprehension. By mastering these techniques, you will be well-equipped to handle a wide range of polynomial inequality problems and apply them to real-world scenarios. The journey through this problem has not only provided a solution but also a deeper appreciation for the elegance and power of algebraic methods in problem-solving.