Problem 39 - Precalculus with Calculus Previews

Question

Find the zeros of the given polynomial function \(f .\) State the multiplicity of each zero. $$ f(x)=x(4 x-5)^{2}(2 x-1)^{3} $$

Step-by-Step Solution

Answer

Zeros: 0 (multiplicity 1), 5/4 (multiplicity 2), 1/2 (multiplicity 3).

Step 1: Understand the problem

The given function is a polynomial: \( f(x) = x(4x-5)^2(2x-1)^3 \). You need to find the zeros of this polynomial and state the multiplicity of each zero.

Step 2: Identify the factors of the polynomial

The function is already factored, consisting of three factors: \( x \), \( (4x-5)^2 \), and \( (2x-1)^3 \). Each factor can contribute to a zero, based on the equation \( f(x) = 0 \).

Step 3: Set each factor equal to zero

To find the zeros, set each factor equal to zero: 1. \( x = 0 \)2. \( 4x - 5 = 0 \)3. \( 2x - 1 = 0 \)

Step 4: Solve each equation for zeros

Solve each equation from the previous step:1. \( x = 0 \) is already solved.2. For \( 4x - 5 = 0 \), add 5 to both sides and then divide by 4. Thus, \( x = \frac{5}{4} \).3. For \( 2x - 1 = 0 \), add 1 to both sides and then divide by 2. Thus, \( x = \frac{1}{2} \).

Step 5: Determine the multiplicity of each zero

The multiplicity refers to the exponent of the factor associated with each zero:1. The zero \( x = 0 \) comes from the factor \( x \), which has an exponent of 1. Therefore, it has a multiplicity of 1.2. The zero \( x = \frac{5}{4} \) comes from the factor \( (4x-5)^2 \), which has an exponent of 2. Therefore, it has a multiplicity of 2.3. The zero \( x = \frac{1}{2} \) comes from the factor \( (2x-1)^3 \), which has an exponent of 3. Therefore, it has a multiplicity of 3.

Key Concepts

Polynomial Factorization

Polynomial factorization involves expressing a polynomial as the product of its simpler polynomial factors. This process makes it easier to identify the roots or zeros of the polynomial. For the polynomial function given in the exercise, we have it factored as:

• \( x \)
• \( (4x - 5)^2 \)
• \( (2x - 1)^3 \)

Each of these expressions is a factor of the polynomial. By setting each factor to zero, you can find the corresponding zeros of the polynomial.
Factoring is particularly useful because it simplifies the process of solving polynomials, as the simpler equation \( f(x) = 0 \) can be solved easily when it is in factorized form. Factorization also highlights the structure and symmetries of the polynomial, helping us identify repeated factors, which lead directly to the concept of multiplicity of zeros.

Multiplicity of Zeros

The multiplicity of a zero refers to the number of times that zero appears as a solution to the polynomial. It is tied to the exponent on the factor associated with each zero. Here’s how it works for our exercise:

• Zero \( x = 0 \) comes from the factor \( x \) and has an exponent of 1, hence a multiplicity of 1.
• Zero \( x = \frac{5}{4} \) comes from \( (4x-5)^2 \) with an exponent of 2, so it has a multiplicity of 2.
• Zero \( x = \frac{1}{2} \) comes from \( (2x-1)^3 \) with an exponent of 3, therefore, it has a multiplicity of 3.

Multiplicity has important implications for the graph of a polynomial function:

• A zero with multiplicity 1 implies that the graph will cross the x-axis at this point.
• A zero with higher multiplicity implies the graph will touch the x-axis and turn around, rather than crossing it. The greater the multiplicity, the "flatter" the graph appears at that zero.

Solving Polynomial Equations

Solving polynomial equations involves finding all of the values for which the polynomial evaluates to zero. These values are known as the polynomial's zeros. In our exercise, the function is already in factored form, which simplifies the task:

• Factor \( x = 0 \) gives the zero \( x = 0 \).
• Solving \( 4x - 5 = 0 \), we add 5 to both sides and divide by 4 to find \( x = \frac{5}{4} \).
• Solving \( 2x - 1 = 0 \), we add 1 to both sides and divide by 2, resulting in \( x = \frac{1}{2} \).

When solving polynomial equations, the goal is always to express the polynomial in a form that makes finding these values straightforward. Factoring the polynomial initially is a critical step, because it transforms the equation into separate, simpler equations. Solving these individual equations will directly yield the zeros, completing the solution process. Understanding this process allows you to apply similar strategies to more complex polynomial equations.