Polynomial Function: Degree 3, Zeros At 5 (Multiplicity 2) & -2

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Hey guys! Today, we're diving deep into the fascinating world of polynomials, specifically focusing on how to construct a polynomial function f(x) of degree 3, given its zeros. This is a fundamental concept in algebra and calculus, and mastering it will definitely level up your math game. We'll walk through the process step-by-step, ensuring you grasp every detail. Let's get started!

Understanding the Basics of Polynomial Zeros

Before we jump into the solution, it’s crucial to understand what polynomial zeros actually are. A zero of a polynomial, sometimes called a root, is a value of x that makes the polynomial equal to zero. In other words, if f(a) = 0, then a is a zero of the polynomial f(x). Zeros tell us a lot about the behavior of the polynomial, particularly where it intersects the x-axis on a graph.

The multiplicity of a zero is the number of times that zero appears as a root of the polynomial. For example, if a polynomial has a factor of (x - a)^2, then a is a zero with multiplicity 2. This multiplicity affects how the graph of the polynomial behaves at that zero. A zero with an even multiplicity (like 2) means the graph touches the x-axis but doesn't cross it, while a zero with an odd multiplicity (like 1 or 3) means the graph crosses the x-axis.

In our problem, we're given that the polynomial f(x) has a zero of 5 with multiplicity 2 and a zero of -2. This means that the factor (x - 5) appears twice in the polynomial, and the factor (x + 2) appears once. Understanding these concepts is crucial for constructing our polynomial.

Building the Polynomial from its Zeros

Now, let's talk about how we can use the zeros to actually build our polynomial. The factor theorem is our best friend here. It states that if a is a zero of the polynomial f(x), then (x - a) is a factor of f(x). Conversely, if (x - a) is a factor of f(x), then a is a zero of f(x). This theorem gives us a direct way to translate zeros into factors.

Since we have a zero of 5 with multiplicity 2, we know that (x - 5)^2 is part of our polynomial. And because we have a zero of -2, we know that (x + 2) is also a factor. So, our polynomial f(x) can be written in the form:

f(x) = k(x - 5)^2(x + 2)

where k is a constant. This constant is important because it scales the polynomial vertically without changing its zeros. To fully define our polynomial, we need to determine a value for k. However, if we are only asked to find a polynomial, we can often assume k = 1 for simplicity. The core structure of the polynomial is determined by its zeros and their multiplicities, which we've already accounted for.

Step-by-Step Solution: Constructing the Polynomial f(x)

Let's break down the process of constructing the polynomial f(x) step-by-step. This will make it super clear and easy to follow. Remember, our goal is to find a polynomial of degree 3 with a zero of 5 (multiplicity 2) and a zero of -2.

Step 1: Identify the Factors

The first step is to identify the factors corresponding to the zeros. We know that:

  • A zero of 5 with multiplicity 2 gives us the factor (x - 5)^2.
  • A zero of -2 gives us the factor (x + 2).

This is a direct application of the factor theorem. We're simply translating the information about the zeros into algebraic expressions that will form our polynomial.

Step 2: Write the General Form of the Polynomial

Next, we combine these factors to write the general form of the polynomial. We include the constant k to account for any vertical scaling:

f(x) = k(x - 5)^2(x + 2)

This equation represents a family of polynomials that all have the given zeros. Each value of k gives us a slightly different polynomial, but they all share the same fundamental shape and zero locations. If we were given an additional point on the polynomial, we could solve for k and find the specific polynomial that passes through that point. However, since we're not given an additional point, we can move on with k as an arbitrary constant.

Step 3: Expand the Polynomial (Optional but Recommended)

To get the polynomial in standard form (i.e., ax^3 + bx^2 + cx + d), we need to expand the expression. This involves multiplying out the factors. Let's start by expanding (x - 5)^2:

(x - 5)^2 = (x - 5)(x - 5) = x^2 - 10x + 25

Now, we multiply this result by (x + 2):

(x^2 - 10x + 25)(x + 2) = x^3 - 10x^2 + 25x + 2x^2 - 20x + 50

Combining like terms, we get:

x^3 - 8x^2 + 5x + 50

So, our polynomial now looks like:

f(x) = k(x^3 - 8x^2 + 5x + 50)

Expanding the polynomial gives us a clearer picture of its coefficients and overall structure. While the factored form is useful for identifying zeros, the expanded form is often more convenient for other algebraic manipulations and for graphing the polynomial.

Step 4: Choose a Value for k (if necessary)

If we're only asked to find a polynomial that satisfies the given conditions, we can choose any non-zero value for k. The simplest choice is often k = 1. This gives us the polynomial:

f(x) = x^3 - 8x^2 + 5x + 50

If we were given an additional constraint, such as the value of f(x) at a particular point, we could solve for k to find the unique polynomial that satisfies all the conditions. For instance, if we knew that f(0) = 100, we could substitute x = 0 into the general form and solve for k. However, in this case, we're only asked to find a polynomial, so k = 1 is perfectly acceptable.

The Final Polynomial

Therefore, a polynomial f(x) of degree 3 that has zeros at 5 (multiplicity 2) and -2 is:

f(x) = x^3 - 8x^2 + 5x + 50

This is just one possible answer, as any non-zero value of k would give us another valid polynomial. However, this is the simplest form, corresponding to k = 1. We've successfully constructed a polynomial from its zeros, which is a fundamental skill in algebra and calculus.

Real-World Applications and Why This Matters

You might be wondering,