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Find a polynomial function with the given real zeros whose graph contains the given point. Zeros: -1 (multiplicity 2), 1 (multiplicity 2) Degree 4 Point: (-2,45)

Short Answer

Expert verified
The polynomial function is f(x) = 5(x^4 - 2x^2 + 1).

Step by step solution

01

Identify the Polynomial Form

Given the zeros -1 (multiplicity 2) and 1 (multiplicity 2), the polynomial can be written as: f(x) = a(x + 1)^2(x - 1)^2
02

Expand the Polynomial

Expand (x + 1)^2(x - 1)^2: (x + 1)^2 = x^2 + 2x + 1 (x - 1)^2 = x^2 - 2x + 1 Using distribution: f(x) = a(x^2 + 2x + 1)(x^2 - 2x + 1)
03

Simplify the Polynomial

Use the distribution to expand and simplify the polynomial: f(x) = a(x^2(x^2 - 2x + 1) + 2x(x^2 - 2x + 1) + 1(x^2 - 2x + 1)) f(x) = a(x^4 - 2x^3 + x^2 + 2x^3 - 4x^2 + 2x + x^2 - 2x + 1) f(x) = a(x^4 - 2x^2 + 1)
04

Determine the Coefficient 'a'

Substitute the point (-2, 45) into the polynomial to find 'a': 45 = a((-2)^4 - 2(-2)^2 + 1) 45 = a(16 - 8 + 1) 45 = a(9) a = 5
05

Write the Final Polynomial

Substitute 'a' back into the polynomial: f(x) = 5(x^2 + 2x + 1)(x^2 - 2x + 1) After expansion and simplification: f(x) = 5(x^4 - 2x^2 + 1)

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Key Concepts

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

real zeros
In polynomial functions, real zeros are the x-values where the polynomial equals zero. These are the points where the graph of the function crosses or touches the x-axis. Identifying the real zeros of a polynomial is essential for sketching its graph and solving equations.

**Key Points to Remember about Real Zeros**
  • Real zeros are found by solving the equation f(x) = 0.
  • Each zero represents an x-value where the polynomial crosses the x-axis.
  • Real zeros can be repeated, impacting the shape of the graph.
In the given exercise, the polynomial function has real zeros at -1 and 1. These zeros indicate that f(x) touches the x-axis at these points.
multiplicity of zeros
Multiplicity of zeros refers to how many times a particular zero appears in a polynomial. It indicates whether the graph of the function merely touches the x-axis at that point or actually crosses it.

**Understanding Multiplicity**
  • A zero with an **odd multiplicity** causes the graph to cross the x-axis.
  • A zero with an **even multiplicity** causes the graph to touch but not cross the x-axis.
  • Higher multiplicity flattens the graph at the zero.
For our exercise, the zeros -1 and 1 both have a multiplicity of 2. This means the graph touches the x-axis at these points but does not cross it.
polynomial expansion
Polynomial expansion involves distributing the factors of a polynomial to write it as a sum of its terms. This process helps in simplifying and solving polynomial equations.

**Steps in Polynomial Expansion**
  • Expand any squared or higher power terms first.
  • Use the distributive property to multiply the expanded terms.
  • Combine like terms to simplify the expression.
In this exercise, we expand \((x + 1)^2(x - 1)^2\) starting with:

\((x + 1)^2 = x^2 + 2x + 1\)
\((x - 1)^2 = x^2 - 2x + 1\)

Using distribution, we get:
\(a(x^2 + 2x + 1)(x^2 - 2x + 1)\)

After expansion and combining like terms, the final polynomial is simplified.
finding coefficients
Finding the coefficients in a polynomial is crucial for writing the exact function. Coefficients are the constants that multiply the variables in each term of the polynomial.

**Process of Finding Coefficients**
  • Form the expanded polynomial expression.
  • Substitute the given point to solve for the leading coefficient.
  • Multiply the polynomial by this coefficient.
In our exercise, we substitute (-2, 45) into the polynomial:

\(45 = a((-2)^4 - 2(-2)^2 + 1) = 9a \).
Solving for 'a' gives us: \ a = 5 \.

Therefore, the final polynomial with the correct coefficient is \ f(x) = 5(x^4 - 2x^2 + 1) \.

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Most popular questions from this chapter

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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