Download 171S4.4q Theorems about Zeros of Polynomial Functions

171S4.4q Theorems about Zeros of Polynomial Functions
MAT 171 Precalculus Algebra
4.4 Theorems about Zeros of Polynomial Functions
Dr. Claude Moore
Cape Fear Community College
CHAPTER 4: Polynomial and Rational Functions
4.1 Polynomial Functions and Models
4.2 Graphing Polynomial Functions
4.3 Polynomial Division; The Remainder and Factor Theorems
4.4 Theorems about Zeros of Polynomial Functions
4.5 Rational Functions
4.6 Polynomial and Rational Inequalities
The following lesson is a brief discussion of and suggestions relative to studying Chapter 4.
171Session4
171Session4 ( Package file )
To view this lesson, click the globe to the lower left.
Use this Excel program to calculate synthetic division problems. cfcc.edu/mathlab/synthetic­division.xlsx
March 26, 2013
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Find a polynomial with specified zeros.
For a polynomial function with integer coefficients,
find the rational zeros and the other zeros, if possible.
Use Descartes’ rule of signs to find information about the number of real zeros of a polynomial function with real coefficients.
The Fundamental Theorem of Algebra
Every polynomial function of degree Click the globe to the left and visit SAS Curriculum Pathways for interactive programs on Synthetic Division. Enter the User Name given by your instructor and use Quick Launch # 2300.
n, with n ≥ 1, has at least one zero in the system of complex numbers.
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Zeros of Polynomial Functions with Real Coefficients
The Fundamental Theorem of Algebra
Example: Find a polynomial function of degree 4 having zeros 1, 2, 4i, and ­4i.
Solution: Such a polynomial has factors (x − 1),(x − 2), (x − 4i), and (x + 4i), so we have: Nonreal Zeros: If a complex number a + bi, b ≠ 0, is a zero of a polynomial function f(x) with real coefficients, then its conjugate, a − bi, is also a zero. (Nonreal zeros occur in conjugate pairs.)
Irrational Zeros: If where a, b, and c are rational and b is not a perfect square, is a zero of a polynomial function f(x) with rational coefficients, then its conjugate is also a zero. Let an = 1:
Example
Suppose that a polynomial function of degree 6 with rational coefficients has ­3 + 2i, ­6i, and as three of its zeros. Find the other zeros.
Solution: The other zeros are the conjugates of the given zeros, ­3 ­ 2i, 6i, and There are no other zeros because the polynomial of degree 6 can have at most 6 zeros.
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Example
Rational Zeros Theorem
Let the polynomial function be represented as
P(x) = anxn + an­1xn­1 + an­2xn­2 + ... + a1x + a0
Given f(x) = 2x3 − 3x2 − 11x + 6:
a) Find the rational zeros and then the other zeros.
b) Factor f(x) into linear factors.
Solution: a) Because the degree of f(x) is 3, there are at most 3 distinct zeros. The possibilities for p/q are:
where all the coefficients are integers. Consider a rational number denoted by p/q, where p and q are relatively prime (having no common factor besides ­1 and 1). If p/q is a zero of P(x), then p is a factor of a0 and q is a factor of an.
Use synthetic division to help determine the zeros. It is easier to consider the integers before the fractions.
We try 1: We try ­1:
Since f(1) = ­6, 1 is Since f(­1) = 12, ­1 is not a zero. not a zero.
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171S4.4q Theorems about Zeros of Polynomial Functions
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Example continued
Descartes’ Rule of Signs
We try 3:
Let P(x) be a polynomial function with real coefficients and a nonzero constant term. Since f(3) = 0, 3 is a zero. Thus x ­ 3 is a factor. Using the results of the division above, we can express f(x) as
f(x) = (x ­ 3)(2x2 + 3x ­ 2) .
We can further factor 2x2 + 3x ­ 2 as (2x ­ 1)(x + 2).
The rational zeros are −2, 3 and
The complete factorization of f(x) is:
The number of positive real zeros of P(x) is either:
1. The same as the number of variations of sign in P(x), or
2. Less than the number of variations of sign in P(x) by a positive even integer.
The number of negative real zeros of P(x) is either:
3. The same as the number of variations of sign in P(­x), or
4. Less than the number of variations of sign in P(­x) by a positive even integer.
A zero of multiplicity m must be counted m times.
f(x) = (2x ­ 1)(x ­ 3)(x + 2)
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Example
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342/2. Find a polynomial function of degree 3 with the given numbers as zeros: ­1, 0, 4
What does Descartes’ rule of signs tell us about the number of positive real zeros and the number of negative real zeros?
There are two variations of sign, so there are either two or zero positive real zeros to the equation.
342/4. Find a polynomial function of degree 3 with the given numbers as zeros: 2, i, ­i
There are two variations of sign, so there are either two or zero negative real zeros to the equation.
Total Number of Zeros (or Roots) = 4:
Possible number of zeros (or roots) by kind:
Positive 2 2 0 0
Negative 2 0 2 0
Nonreal 0 2 2 4
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342/6. Find a polynomial function of degree 3 with the given numbers as zeros: ­5, √3, ­ √3
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342/14. Find a polynomial function of degree 4 with ­ 2 as a zero of multiplicity 1, 3 as a zero of multiplicity 2, and ­1 as a zero of multiplicity 1. 342/8. Find a polynomial function of degree 3 with the given numbers as zeros: ­4, 1 ­ √5, 1 + √5
342/16. Find a polynomial function of degree 5 with ­1/2 as a zero of multiplicity 2, 0 as a zero of multiplicity 1, and 1 as a zero of multiplicity 2. Oct 25­1:26 PM
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171S4.4q Theorems about Zeros of Polynomial Functions
342/14. Find a polynomial function of degree 4 with ­ 2 as a zero of multiplicity 1, 3 as a zero of multiplicity 2, and ­1 as a zero of multiplicity 1. March 26, 2013
343/24. Suppose that a polynomial function of degree 4 with rational coefficients has the given numbers as zeros. Find the other zero(s): 6 ­ 5i, ­1 + √7
n = 4 means that the polynomial function has 4 roots (zeros) when we include complex solutions (roots or zeros).
x = 6 ­ 5i and ­1 + √7 are given as two roots (zeros).
Since x = 6 ­ 5i is a root, we know that the conjugate x = 6 + 5i is a root.
342/16. Find a polynomial function of degree 5 with ­1/2 as a zero of multiplicity 2, 0 as a zero of multiplicity 1, and 1 as a zero of multiplicity 2. Since x = ­1 + √7 is a root, we know that the conjugate x = ­1 ­ √7 is a root.
Therefore, the four (4) roots (zeros) are x = 6 ­ 5i, 6 + 5i, x = ­1 + √7, and x = ­1 ­ √7.
If you want to write the polynomial function, use the roots (zeros) to obtain the factors. Multiply the factors to get the polynomial function as shown below.
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343/26. Suppose that a polynomial function of degree 5 with rational coefficients has the given numbers as zeros. Find the other zero(s): 3/4, ­ √3, 2i Oct 25­1:26 PM
343/29. Suppose that a polynomial function of degree 5 with rational coefficients has the given numbers as zeros. Find the other zero(s): 6, ­3 + 4i, 4 ­ √5 n = 5 means that the polynomial function has 5 roots (zeros) when we include complex solutions (roots or zeros).
x = 3/4, x = ­√3, and x = 2i are given as three roots (zeros).
x = 6, x = ­3 + 4i, and x = 4 ­ √5 are given as three roots (zeros).
Since x = ­3 + 4i is a root, we know that the conjugate x = ­3 ­ 4i is a root.
Since x = ­√3 is a root, we know that the conjugate Since x = 4 ­ √5 is a root, we know that the conjugate x = √3 is a root.
Since x = 2i is a root, we know that the conjugate x = ­2i is a root.
Therefore, the five (5) roots (zeros) are x = 4 + √5 is a root.
Therefore, the five (5) roots (zeros) are n = 5 means that the polynomial function has 5 roots (zeros) when we include complex solutions (roots or zeros).
x = 3/4, x = ­√3, x = ­√3, x = 2i, and x = ­2i.
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343/36. Find a polynomial function of lowest degree with rational coefficients that has the given numbers as some of its zeros: ­5i
x = 6, x = ­3 + 4i, x = ­3 ­ 4i, x = 4 ­ √5, and x = 4 + √5.
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343/44. Given that the polynomial function has the given zero, find the other zeros: f(x) = x3 ­ x2 + x ­ 1; 1
343/46. Given that the polynomial function has the given zero, find the other zeros: f(x) = x4 ­ 16; 2i
Find the polynomial function with lowest degree (smallest n) that has x = ­5i as a root (zero). The conjugate x = 5i is also a root (zero) of the function.
Thus, we can write f(x) = (x + 5i )(x ­ 5i ) = x2 + 25.
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171S4.4q Theorems about Zeros of Polynomial Functions
343/47. Given that the polynomial function has the given zero, find the other zeros: f(x) = x3 ­ 6x2 + 13x ­ 20; 4 March 26, 2013
343/50. List all possible rational zeros of the function: f(x) = x7 + 37x5 ­ 6x2 + 12 343/48. Given that the polynomial function has the given zero, find the other zeros: f(x) = x3 ­ 8; 2 343/52. List all possible rational zeros of the function: f(x) = 3x3 ­ x2 + 6x ­ 9
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343/53. List all possible rational zeros of the function: f(x) = 15x6 + 47x2 + 2 Oct 25­1:26 PM
343/62. For each polynomial function: a) Find the rational zeros and then the other zeros; that is, solve f(x) = 0. b) Factor f(x) into linear factors. f(x) = 2x3 + 7x2 + 2x ­ 8
343/54. List all possible rational zeros of the function: f(x) = 10x25 + 3x17 ­ 35x + 6
Use this Excel program to calculate synthetic division problems. cfcc.edu/mathlab/synthetic­division.xlsx
Other roots are (­3 ­ √41) / 4 and (­3 + √41) / 4.
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343/64. For each polynomial function: a) Find the rational zeros and then the other zeros; that is, solve f(x) = 0. b) Factor f(x) into linear factors. f(x) = 3x4 ­ 4x3 + x2 + 6x ­ 2
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343/66. For each polynomial function: a) Find the rational zeros and then the other zeros; that is, solve f(x) = 0. b) Factor f(x) into linear factors. f(x) = x4 + 5x3 ­ 27x2 + 31x ­ 10 Use this Excel program to calculate synthetic division problems. cfcc.edu/mathlab/synthetic­division.xlsx
Use this Excel program to calculate synthetic division problems. cfcc.edu/mathlab/synthetic­division.xlsx
Divide 3x2 ­ 6x + 6 = 0 by ­3 to yield x2 ­ 2x + 2 = 0. This gives x = 1+ i and x = 1­ i to go with x = ­1 and x = 1/3.
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Divide 3x2 ­ 6x + 6 = 0 by ­3 to yield x2 ­ 2x + 2 = 0. This gives x = 1+ i and x = 1­ i to go with x = ­1 and x = 1/3.
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171S4.4q Theorems about Zeros of Polynomial Functions
343/71. For each polynomial function: a) Find the rational zeros and then the other zeros; that is, solve f(x) = 0. b) Factor f(x) into linear factors. f(x) = (1/3)x3 ­ (1/2)x2 ­ (1/6)x + 1/6
March 26, 2013
344/74. Find only the rational zeros of the function.
f(x) = x4 ­ 3x3 ­ 9x2 ­ 3x ­ 10 Use this Excel program to calculate synthetic division problems. cfcc.edu/mathlab/synthetic­division.xlsx
344/76. Find only the rational zeros of the function.
f(x) = 2x3 + 3x2 + 2x + 3
Use this Excel program to calculate synthetic division problems. cfcc.edu/mathlab/synthetic­division.xlsx
Use this Excel program to calculate synthetic division problems. cfcc.edu/mathlab/synthetic­division.xlsx
The quadratic (1/3)x2 ­ (1/3)x ­ 1/3 = 0 multiplied by 3 yields an equivalent equation x2 ­ x ­ 1 = 0. This gives x = (1+ √5) / 2 and x = (1­ √5) / 2 to go with x = 1/2.
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344/92. What does Descartes rule of signs tell you about the number of positive real zeros and the number of negative real zeros of the function?
Q(x) = x4 ­ 2x2 + 12x ­ 8
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344/98. Sketch the graph of the polynomial function. Follow the procedure outlined on p. 317. Use the rational zeros theorem when finding the zeros.
f(x) = 3x3 ­ 4x2 ­ 5x + 2 344/96. What does Descartes rule of signs tell you about the number of positive real zeros and the number of negative real zeros of the function?
f(x) = x4 ­ 9x2 ­ 6x + 4
f(x) = +1x4 - 9x2 - 6x + 4 has two (2) sign changes.
Thus, there are two or zero positive real roots (or
zeros).
f(-x) = +1x4 - 9x2 + 6x + 4 has two (2) sign changes.
Thus, there are two or zero negative real roots (or
zeros).
Oct 25­1:26 PM
344/100. Sketch the graph of the polynomial function. Follow the procedure outlined on p. 317. Use the rational zeros theorem when finding the zeros.
f(x) = 4x4 ­ 37x2 + 9 Oct 25­1:26 PM
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344/98. Sketch the graph of the polynomial function. Follow the procedure outlined on p. 317. Use the rational zeros theorem when finding the zeros.
f(x) = 3x3 ­ 4x2 ­ 5x + 2 Oct 25­1:26 PM
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171S4.4q Theorems about Zeros of Polynomial Functions
March 26, 2013
344/100. Sketch the graph of the polynomial function. Follow the procedure outlined on p. 317. Use the rational zeros theorem when finding the zeros.
f(x) = 4x4 ­ 37x2 + 9 Oct 25­1:26 PM
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