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
MAT 171 Precalculus Algebra
Dr. Claude Moore Cape Fear Community College
See the following lesson in Course Documents of CourseCompass:
171Session4
171Session4 ( Package file )
This lesson is a brief discussion of and suggestions relative to studying Chapter 4.
4.4 Theorems about Zeros of Polynomial Functions
•
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 n, with n ≥ 1, has at least one zero in the system of complex numbers.
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:
Let an = 1:
Zeros of Polynomial Functions with Real Coefficients 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.
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.
Rational Zeros Theorem
Let
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 a
0and q is a factor of a
n.
Example
Given f(x) = 2x
3− 3x
2− 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:
Example continued
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.
Example continued
We try 3:
.
We can further factor 2x2 + 3x 2 as (2x 1)(x + 2).
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
Example continued
The rational zeros are −2, 3 and The complete factorization of f(x) is:
Descartes’ Rule of Signs
Let P(x) be a polynomial function with real coefficients and a nonzero constant term. 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
Example
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.
Example continued
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
339/4. Find a polynomial function of degree 3 with the given numbers as zeros: 2, i, i
n = 3; x = 2, x = i, x = i f(x) = (x 2)(x i)(x + i) = (x 2)(x2 + 1) f(x) = x3 2x2 + x 2
339/8. Find a polynomial function of degree 3 with the given
numbers as zeros: 4, 1 √5, 1 + √5
zero of multiplicity 1, 3 as a zero of multiplicity 2, and 1 as a
zero of multiplicity 1. 340/24. 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
We only needed to find the other roots or zeros. We did not need to find the polynomial function.
340/29. 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 = 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 = 4 √5 is a root, we know that the conjugate x = 4 + √5 is a root.
340/36. Find a polynomial function of lowest degree with rational coefficients that has the given numbers as some of its zeros:
5i
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.
340/46. Given that the polynomial function has the given zero, find the other zeros: f(x) = x
4 16; 2i
340/52. List all possible rational zeros of the function:
f(x) = 3x
3 x
2+ 6x 9
340/62. 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) = 3x
4 4x
3+ x
2+ 6x 2
NOTE: The original problem was copied wrong. The roots are x = 1, x = 1/3, x = 2i, and x = 2i.
340/69. 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)x
3 (1/2)x
2 (1/6)x + 1/6
340/76. Find only the rational zeros of the function.
f(x) = x
4+ 6x
3+ 17x
2+ 36x + 66
q ∈ {±1} p ∈ {±1, 2, 3, 6, 11, 22, 33, 66}
I tried all 16 possible rational roots and found that none of them worked. So, f(x) has NO rational roots.
