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Challenge Problem Find the real zeros of $$ f(x)=3\left(x^{2}-1\right)\left(x^{2}+4 x+3\right)^{2} $$ and their multiplicity.

Short Answer

Expert verified
The zeros are \( x = 1 \) (multiplicity 1), \( x = -1 \) (multiplicity 3), and \( x = -3 \) (multiplicity 2).

Step by step solution

01

- Factorize Each Polynomial

First, we need to factorize the polynomials within the function if possible. For the terms in the function:1. Factorize \( x^{2} - 1 \) as follows: \[ x^{2} - 1 = (x - 1)(x + 1) \].2. Factorize \( x^{2} + 4x + 3 \) as follows: \[ x^{2} + 4x + 3 = (x + 1)(x + 3) \].
02

- Rewrite the Function

Substitute the factors found in Step 1 into the function. So, \( f(x) = 3 (x - 1)(x + 1)(x + 1)^{2}(x + 3)^{2} \).
03

- Set Function to Zero

To find the zeros of \( f(x) \), set the function equal to zero and solve for \( x \).Thus, set each factor equal to zero:1. \( x - 1 = 0 \)2. \( x + 1 = 0 \)3. \( x + 3 = 0 \)
04

- Solve for Zeros

Solve each equation from Step 3:1. \( x = 1 \) for \( x - 1 = 0 \)2. \( x = -1 \) for \( x + 1 = 0 \)3. \( x = -3 \) for \( x + 3 = 0 \)
05

- Determine Multiplicities

From the rewritten function \( f(x) = 3 (x - 1)(x + 1)(x + 1)^{2}(x + 3)^{2} \), determine the multiplicities of each zero:1. \( x = 1 \) has a multiplicity of 1 (since it appears once).2. \( x = -1 \) has a multiplicity of 3 (since it appears once as \( x + 1 \) and twice as \( (x + 1)^{2} \).3. \( x = -3 \) has a multiplicity of 2 (since it appears twice as \( (x + 3)^{2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factoring Polynomials
When solving polynomial equations, one of the first steps is often factoring the polynomial. Factoring means breaking down a polynomial into simpler expressions called factors that, when multiplied together, give the original polynomial. For example, the polynomial \( x^2 - 1 \) can be factored as \( (x - 1)(x + 1) \), because multiplying \( (x - 1) \) and \( (x + 1) \) returns \( x^2 - 1 \). Another polynomial, \( x^2 + 4x + 3 \), factors to \( (x + 1)(x + 3) \). Each factor like \( x - 1 \) or \( x + 3 \) can further reveal important properties about the polynomial, such as its zeros and their multiplicities. Understanding how to factor polynomials is crucial for solving polynomial equations and finding real zeros efficiently.
Multiplicity of Zeros
Multiplicity refers to the number of times a particular zero appears as a root of the polynomial. In the given example, we had to find the zeros of the function \( f(x) = 3 (x - 1)(x + 1)(x + 1)^2 (x + 3)^2 \). Each factor contributes to the final count of zeros. For instance, the zero \( x = 1 \) appears only once, giving it a multiplicity of 1. The zero \( x = -1 \) shows up three times: once as \( x + 1 \) and twice as \( (x + 1)^2 \), thus it has a multiplicity of 3. Similarly, \( x = -3 \) is counted twice due to the \( (x + 3)^2 \) term, making its multiplicity 2. Recognizing multiplicities is vital because it affects the behavior of the polynomial function at those points.
Solving Polynomial Equations
To solve a polynomial equation, we need to set the polynomial equal to zero. This reveals where the polynomial crosses the x-axis, or its roots/zeros. From the previous steps, we have the function \( f(x) = 3 (x - 1)(x + 1)(x + 1)^2 (x + 3)^2 \). To find the zeros, we set the function equal to zero and solve: \[ 3 (x - 1)(x + 1)(x + 1)^2 (x + 3)^2 = 0 \]. This equation implies we need to solve each factor for zero: \( x - 1 = 0 \), \( x + 1 = 0 \), and \( x + 3 = 0 \). Solving these gives \( x = 1 \), \( x = -1 \), and \( x = -3 \), which are the roots of the polynomial. Solving polynomial equations often involves factoring, setting each factor to zero, and solving these simpler equations.
Real Zeros of Functions
Real zeros of a function are the values of \( x \) which make the function equal to zero. These are the x-values where the graph of the function touches or crosses the x-axis. For our example function \( f(x) = 3 (x - 1)(x + 1)(x + 1)^2 (x + 3)^2 \), the real zeros can be found by solving \( 3 (x - 1)(x + 1)(x + 1)^2 (x + 3)^2 = 0 \). This reveals that the real zeros of the function are \( x = 1 \), \( x = -1 \), and \( x = -3 \). The graph of this function touches the x-axis at these points. Finding real zeros is essential, as these points provide key information about the behavior and roots of the polynomial equation.

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Most popular questions from this chapter

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. $$ \text { Solve: } \frac{3 x}{3 x+1}=\frac{x-2}{x+5} $$

Suppose that the daily cost \(C\) of manufacturing bicycles is given by \(C(x)=80 x+5000 .\) Then the average daily cost \(\bar{C}\) is given by \(\bar{C}(x)=\frac{80 x+5000}{x} .\) How many bicycles must be produced each day for the average cost to be no more than \(\$ 100 ?\)

Solve each equation in the real number system. $$ x^{4}-x^{3}+2 x^{2}-4 x-8=0 $$

Use the Rational Zeros Theorem to find all the real zeros of each polynomial function. Use the zeros to factor \(f\) over the real numbers. $$ f(x)=2 x^{3}+x^{2}+2 x+1 $$

We begin with two consecutive integers, \(a\) and \(a+1,\) for which \(f(a)\) and \(f(a+1)\) are of opposite sign. Evaluate \(f\) at the midpoint \(m_{1}\) of \(a\) and \(a+1 .\) If \(f\left(m_{1}\right)=0,\) then \(m_{1}\) is the zero of \(f,\) and we are finished. Otherwise, \(f\left(m_{1}\right)\) is of opposite sign to either \(f(a)\) or \(f(a+1) .\) Suppose that it is \(f(a)\) and \(f\left(m_{1}\right)\) that are of opposite sign. Now evaluate \(f\) at the midpoint \(m_{2}\) of \(a\) and \(m_{1} .\) Repeat this process until the desired degree of accuracy is obtained. Note that each iteration places the zero in an interval whose length is half that of the previous interval. Use the bisection method to approximate the zero of \(f(x)=8 x^{4}-2 x^{2}+5 x-1\) in the interval [0,1] correct to three decimal places. [Hint: The process ends when both endpoints agree to the desired number of decimal places.

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