Assignment: Objectives: P. 179: 9 To find the zeros of a

Objectives:
1. To find the zeros of a
polynomial using
one or more of our 7
handy zero-finding
tools
•
•
•
•
•
•
Assignment:
P. 179: 9
P. 179: 11-20 (Some)
P. 180: 37-42 (Some)
P. 180: 47-54 (Some)
P. 180: 59-72 (Some)
P. 181: 79-86 (Some)
– And imaginary
• P. 182: 107
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?
• 1777-1855
• German
mathematician/Child
prodigy
• Could add the
numbers 1 to 100
really fast, even as a
kid!
(Another avid stamp collector)
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.
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.
• Perhaps more useful than this Fun Theorem is
its 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.
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.
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.
• When a solution is repeated k times, that
solution is said to be a repeated root with a
multiplicity of k.
The graphs below are of second degree
polynomials.
2 zeros
1 zero, repeated
x2
0 real zeros = 2
imaginary
The graphs below are of third degree
polynomials.
1 zero,
3 zeros
2 zeros,
1 repeated x2
2 imaginary
The graphs below are of fourth degree
polynomials.
The problem with the Fundamental Theorem of
Algebra and its Corollary is that, while quite
informative, they only tell us that a certain
number of zeros exist; however, they don’t tell
us how to find them.
So what we need is a tool box (or bag or belt—
depends on how you store your tools) that
helps us find those pesky zeros.
You will be able
to use a variety
of tools to find
the zeros of a
polynomial
function
Find the other zeros of f(x) = 10x3 – 81x2 + 71x + 42
given that f(7) = 0.
Find the other zeros of f(x) = 10x3 – 81x2 + 71x + 42
given that f(7) = 0.
Writing each zero as a rational number, we have:
Factors of 10,
the leading
coefficient.
7 3
2
,
, and 
1 2
5
Factors of 42,
the constant
term.
Notice that the numerators are factors of 42 and the
denominators are factors of 10
If f(x) = anxn + … + a1x + a0 has integer
coefficients, then every rational zero of f(x)
has the following form:
factor of constant term a0
p

q factor of leading coefficient an
Important note: These factors can be either
positive or negative.
Here’s another way to think about The Rational
Zero Test. Consider the function below:
f ( x)  6 x3  x 2  47 x  30
In factored form:
f ( x)  (2 x  5)( x  3)(3x  2)
And here are the zeros:
5
2
x   , 3, and 
2
3
Now work backwards from the factors: What
would be the constant term and the leading
coefficient?
f ( x)  6 x3  x 2  47 x  30
f ( x)  (2 x  5)( x  3)(3x  2)
(5)(3)(2)  30
(2 x)( x)(3x)  6 x3
Factors of 6
5
2
x   , 3, and 
2
3
Factors of −30
To help you remember the order of the factors when
using the Rational Zero Theorem, consider the linear
function:
y  6 x  10
To find the zero(s), let y = 0 and solve for x.
0  6x 10
Factor of 6,
the leading
coefficient.
10  6x
5
x
3
Factor of -10,
the constant
term.
When solving this equation, first you add/subtract
the constant term to the opposite side of the
equation, and then you divide by the leading
coefficient.
0  6x 10
Factor of 6,
the leading
coefficient.
10  6x
5
x
3
Factor of -10,
the constant
term.
List the possible rational zeros of f(x) using the
Rational Zero Theorem.
1. f ( x)  x3  x 2  3x  6 2. f ( x)  2 x3  7 x 2  4 x  8
The Rational Zero Test gives us a list of possible
rational zeros, some of them work, lots of
them do not.
To verify if a number qualifies as a zero, use
synthetic division. Remember, we know a number
is a zero of a function if when we divide, we get a
remainder of 0.
– Once you find a zero, you’ve reduced the degree of your
polynomial. Now it’s easier to solve!
Find all real zeros of the function.
f ( x)  x3  4 x 2  15x  18
Find all real zeros of the function.
𝑓 𝑥 = 24𝑥 3 − 10𝑥 2 − 41𝑥 − 20
Find all real zeros of the function.
𝑓 𝑥 = 24𝑥 3 − 10𝑥 2 − 41𝑥 − 20
Notice that when the leading coefficient is not 1, the number of
possible rational zeros increases in an unsavory way. Surely
there has to be a way to narrow the possibilities… We’ll get to
one of those tools shortly.
Perhaps you’ve noticed this while repeatedly
using the quadratic formula: Irrational and
Complex solutions come in conjugate pairs.
Let f be a polynomial with real coefficients:
If 𝑎 + 𝑏 is a zero of f,
then so is 𝑎 − 𝑏
If 𝑎 + 𝑏𝑖 is a zero of f,
then so is 𝑎 − 𝑏𝑖
(These can only be found with the Quadratic
Formula or Completing the Square)
Use the previous Conjugates Tool to explain why
a polynomial function (with real coefficients)
of odd degree must always have at least one
real root.
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
Given that 2i is a zero of the function below, find
all other zeros.
f ( x)  x 4  4 x3  8x 2  16 x  16
This one seems obvious, right? There’s a
theorem that states:
You can write a polynomial function of
degree n as a product of n linear factors:
𝑓 𝑥 = 𝑥 − 𝑐1 𝑥 − 𝑐2 𝑥 − 𝑐3 ⋯ 𝑥 − 𝑐𝑛
where ci are complex numbers.
This means that if we’re given a factor or if we
stumble upon a factor, we divide that out and
try to factor the rest of the polynomial.
However, if our polynomial has imaginary or
irrational factors, factoring is next to
impossible.
For those, we need the quadratic formula or
completing the square.
Once we find enough factors to be left with a
quadratic polynomial, we can complete the
square or use the quadratic formula to find
the two remaining zeros.
Factor (if possible)
𝑝(𝑥)
Polynomial
Function
Factor:
RZT, Division, etc.
Complete the Square
Quadratic Formula
𝑝 𝑥 = 𝑥 − 𝑐1 𝑥 − 𝑐2 ⋯ 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
Once we find enough factors to be left with a
quadratic polynomial, we can complete the
square or use the quadratic formula to find
the two remaining zeros.
2
0

ax
 bx  c , then
If
b  b  4ac
x
2a
2
Write h(x) = x3 – 11x2 + 41x – 51 as the product
of linear factors, and list all of its zeros.
The Rational Zero Test can give us a seemingly
endless list of possible zeros, but which one
do we test first?
We could cheat a bit by looking at a graph
of the function to see where it intercepted
the x-axis.
Find all real zeros of the function.
f ( x)  2 x3  5 x 2  11x  14
Another way to narrow
down our rational zero
choices is by using the
Location Principle.
Consider the table of
values. What can you
conclude about the
location of one of the
zeros of f(x)?
x
0
1
2
3
4
5
f(x)
-6
-12
28
150
390
784
Location Principle
If f is a polynomial
function and a and
b are two numbers
such that f(a) < 0
and f(b) > 0, then f
has at least one zero
between a and b.
• For example, if
f(2) = −3 and f(3) = 5,
then there must be a
zero between x = 2 and
x = 3. This is because a
continuous curve
would cross the x-axis
when connecting these
points.
Location Principle
If f is a polynomial
function and a and
b are two numbers
such that f(a) < 0
and f(b) > 0, then f
has at least one zero
between a and b.
There must
be a zero
between 2
and 3 since
the graph
must cross
the 𝑥-axis to
connect
these
points.
Location Principle
If f is a polynomial
function and a and
b are two numbers
such that f(a) < 0
and f(b) > 0, then f
has at least one zero
between a and b.
• So we could use a
table of values with
the Location Principle
to narrow our rational
zero choices. Just look
for where the
function-values change
signs: + to – or – to +.
Find all real zeros of the function.
f ( x)  2 x3  5 x 2  11x  14
x
y
-4
-18
-3
10
-2
12
-1
0
0
-14
1
-18
2
0
3
52
The final tool in our handy box/bag/belt is called
Descartes Rule of Signs. Based just on the
variation in signs (from + to − or − to +) in f (x)
and f (−x), we can determine the possible
number of positive and negative real zeros.
But first, here’s another stamp…
•
•
•
•
(Really, another stamp?)
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
Let f(x) = anxn + an – 1xn – 1 + … + a2x2 + a1x + a0 be a
polynomial with real coefficients.
1. 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.
2. 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 than this by an even number.
Use Descartes Rule of
Signs to determine
the possible number
of positive real zeros,
negative real zeros
and imaginary zeros
for
f(x) = 2x6 – 3x2 – x + 1.
f(x) =
2x6
f(−x) =
3x2
–
2x6
–
–x+1
3x2
2(+)
or
0(+)
+x+1
2(−)
or
0(−)
Total
(+)
(−)
𝒊
6
2
2
0
0
2
0
2
0
2
4
4
6
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.
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
Objectives:
1. To find the zeros of a
polynomial using
one or more of our 7
handy zero-finding
tools
•
•
•
•
•
•
Assignment:
P. 179: 9
P. 179: 11-20 (Some)
P. 180: 37-42 (Some)
P. 180: 47-54 (Some)
P. 180: 59-72 (Some)
P. 181: 79-86 (Some)
– And imaginary
• P. 182: 107