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5.7: Fundamental Theorem of Algebra Assignment: P. 383-386: 1, 2, 3-33 M3,

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5.7: Fundamental Theorem of Algebra Assignment: P. 383-386: 1, 2, 3-33 M3,
5.7: Fundamental Theorem of Algebra
1.
2.
3.
4.
Objectives:
To apply the Fundamental
Theorem of Algebra and its
Corollary
To determine the behavior
of the graph of a function
near its zeros
To apply the Conjugates
Theorems
To use Descartes Rule of
Signs to determine the
number of +/- real zeros
Assignment:
• P. 383-386: 1, 2, 3-33 M3,
34-40 even, 50, 53-56,
58, 65-67
• Descartes and
Reflections: 1-4, 5-15 odd
• Graphical Behavior Near
Zeros: 1, 3-6
Vocabulary
Theorem
Corollary
Real Numbers
Complex Numbers
Irrational Conjugates
Complex Conjugates
You will be able to apply the Fundamental
Theorem of Algebra and it’s Corollary
Objective 1
Exercise 1
1. How many solutions does the equation
x4 + 8x2 – 5x + 2 = 0 have?
2. How many zeros does the function
f(x) = x3 + x2 – 3x – 3 have?
Exercise 2
Determine the degree of the polynomial functions.
How many zeros does each have? Use a
graphing calculator to graph each function. Find
all the zeros of each function.
1. f (x) = x3 – x2 – 25x + 25
2. g(x) = x3 + 7x2 + 12x + 10
Exercise 3
Determine the degree of the polynomial functions.
How many zeros does each have? Use a
graphing calculator to graph each function. How
many x-intercepts does each have?
1. f (x) = x3 – x2 – 25x + 25
2. g(x) = x3 + 7x2 + 12x + 10
Degrees and Zeros
The graphs below are of second degree
polynomials.
2 zeros
1 zero,
repeated x2
0 real zeros =
2 imaginary
Degrees and Zeros
The graphs below are of third degree
polynomials.
3 zeros
2 zeros,
1 repeated x2
1 zero,
2 imaginary
Degrees and Zeros
The graphs below are of fourth degree
polynomials.
Exercise 4
1. How many zeros does a quintic
polynomial have?
2. How many x-intercepts does a quintic
polynomial have?
3. Why are the answers above not
necessarily the same?
Karl Fredrich Gauss
• 1777-1855
• German
mathematician/Child
prodigy
• Could add the
numbers 1 to 100
really fast, even as a
kid!
Fun Theorem
Fundamental Theorem of Algebra
If f(x) is a polynomial function of degree n,
where n > 0, then f(x) = 0 has at least one
solution in the set of complex numbers.
This means that a
polynomial function has at
least one complex zero
Also, since real numbers
are complex numbers,
the solution could be a
real one
Fun Theorem
Fundamental Theorem of Algebra
If f(x) is a polynomial function of degree n,
where n > 0, then f(x) = 0 has at least one
solution in the set of complex numbers.
– First proven by Gauss after several unsuccessful
attempts by numerous, famous mathematicians
– Perhaps more useful than this Fun Theorem is its
Corollary
Fun Corollary
Fundamental Theorem of Algebra Corollary
If f(x) is a polynomial function of degree n, then
f(x) = 0 has exactly n solutions provided that
each solution repeated k times is counted as k
solutions.
This means that an
nth degree
polynomial has n
solutions
• When a solution is repeated
k times, that solution is said
to be a repeated root with a
multiplicity of k.
Fun Theorem (Lite)
A consequent of the
Fundamental Theorem of
Algebra and it’s Corollary is
that:
An 𝑛th degree polynomial
has 𝑛 zeros.
Sometimes at least one of these zeros
repeats 𝑘 times and is said to be a
repeated root with a multiplicity of 𝑘.
y  x2 ( x  3)( x  5)3
x0
x0
x3
x  5
x  5
x  5
6 total zeros, but only 3
x-intercepts (some repeat)
You will be able to
determine the behavior
of the graph of a function
near its zeros
Exercise 5
Determine the degree of the following polynomial
functions. How many zeros does each have?
Use a graphing calculator to determine how the
multiplicity of each zero affects the graph.
1. f(x) = (x + 5)(x – 1)2
2. g(x) = (x + 5)3(x – 1)4
3. h(x) = (x + 5)5(x – 1)6
Behavior Near Zeros
Real Zeros:
• Only real zeros are x-intercepts.
Imaginary zeros do not touch
the x-axis.
Odd Multiplicity:
• A zero of odd multiplicity
crosses the x-axis at that zero.
Even Multiplicity:
• A zero of even multiplicity is
tangent to the x-axis at that
zero.
Odd
Multiplicity
Even
Multiplicity
 x  1
 x  2
3
2
You will be able to apply the
Conjugates Theorems
Exercise 6
Find all the zeros of each function. Is there any
relationship between the zeros?
1. f(x) = x3 + 3x2 – 14x – 20
2. g(x) = x3 + 3x2 + 16x + 130
Conjugates Theorem 1
Complex Conjugates
Theorem
If 𝑓 is a polynomial
function with real
coefficients, and
𝑎 + 𝑏𝑖 is a zero of
𝑓, then 𝑎– 𝑏𝑖 is
also a zero of f.
• This means imaginary
solutions always come
in conjugate pairs.
• You have to use the
Quadratic formula (or
Completing the
Square) to find them.
Conjugates Theorem 1
Complex Conjugates
Theorem
If 𝑓 is a polynomial
function with real
coefficients, and
𝑎 + 𝑏𝑖 is a zero of
𝑓, then 𝑎– 𝑏𝑖 is
also a zero of f.
This means imaginary
solutions always
come in conjugate
pairs.
• You have to use the
Quadratic formula (or
Completing the
Square) to find them.
Conjugates Theorem 1I
Irrational Conjugates
Theorem
If 𝑓 is a polynomial
function with real
coefficients, and 𝑎 and 𝑏
are rational numbers
such that 𝑎 + 𝑏 is an
irrational zero of 𝑓, then
𝑎– 𝑏 is also a zero of 𝑓.
• These also must
be found using the
Quadratic Formula
(or by Completing
the Square).
This means
irrational solutions
always come in
conjugate pairs.
Exercise 7
Use the Complex Conjugates Theorem to
explain why a polynomial function (with
real coefficients) of odd degree must
always have at least one real root.
Exercise 8
Write a polynomial function f of least degree that
has rational coefficients, a leading coefficient of
1, and 2 and −2 – 5i as zeros.
x  2  5i
x2
x  2  5i
x2  0
2
2
 x  2   5i 
x2  4 x  4  25i 2  25
x2  4 x  29  0
f ( x)   x  2  x 2  4 x  29 
f ( x)  x3  2 x 2  21x  58
Exercise 9
Write a polynomial function f of least degree
that has rational coefficients, a leading
coefficient of 1, and the given zeros.
1. −1, 2, 4
2. 4, 1 + √5
3. 2, 2i, 4 – √6
Objective 4
You will be able
to use Descartes
Rule of Signs to
determine the
number of +/real zeros
Rene Descartes
•
•
•
•
1596-1650
French philosopher-etc.
Cogito Ergo Sum
A fly taught him about
the Cartesian
coordinate plane and
analytic geometry, for
which he took full credit
Descartes Rule of Signs
Let f(x) = anxn + an – 1xn – 1 + … + a2x2 + a1x + a0 be a
polynomial with real coefficients.
The number of positive real zeros of 𝑓 is equal to
the number of changes in sign of the coefficients
of 𝑓(𝑥) or is less that this by an even number.
The number of negative real zeros of 𝑓 is equal to
the number of changes in sign of the coefficients
of 𝑓(−𝑥) or is less that this by an even number.
Exercise 10
Use Descartes Rule of Signs to determine the
possible number of positive real zeros, negative
real zeros and imaginary zeros for the
polynomial shown.
f ( x)  2 x 6  3 x 2  x  1
Total
(+)
(−)
i
6
2
2
0
0
2
0
2
0
2
4
4
6
2 (+) or 0 (+)
f (  x)  2 x 6  3 x 2  x  1
2 (−) or 0 (−)
Total − [ (+) + (−) ] = i
Following Descartes Rules
Use Descartes Rule of Signs to determine the possible
types of zeros:
1. Use the degree to determine the total number of zeros.
2. Count the number of sign changes in 𝑓(𝑥). This is the
possible # of (+) real zeros—or less by an even #.
3. Change the signs of the odd-powered variables, and
then count the sign changes. This is the possible # of
(−) real zeros—or less by an even #.
4. Use a table to pair up the possible (+) with the possible
(−). Subtract these from the total. This must be the
possible imaginary zeros.
Exercise 11
Use Descartes Rule of Signs to determine the
possible number of positive real zeros, negative
real zeros and imaginary zeros for each function.
1. f(x) = x3 + 2x – 11
2. g(x) = 2x4 – 8x3 + 6x2 – 3x + 1
5.7: Fundamental Theorem of Algebra
1.
2.
3.
4.
Objectives:
To apply the Fundamental
Theorem of Algebra and its
Corollary
To determine the behavior
of the graph of a function
near its zeros
To apply the Conjugates
Theorems
To use Descartes Rule of
Signs to determine the
number of +/- real zeros
Assignment
• P. 383-386: 1, 2, 3-33
M3, 34-40 even, 50,
53-56, 58, 65-67
• Descartes and
Reflections: 1-4, 5-15
odd
• Graphical Behavior
Near Zeros: 1, 3-6
“The sum of the numbers 1 to 100 is 5050!”
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