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8/4/2019 Obj. 6 Zeros of Polynomial Functions (Presentation)
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Obj. 6 Zeros of Polynomial
Functions
Unit 2 Quadratic and Polynomial Functions
8/4/2019 Obj. 6 Zeros of Polynomial Functions (Presentation)
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Concepts and Objectives
Zeros of Polynomial Functions (Obj. #6)
Find rational zeros of a polynomial function Use the Fundamental Theorem of Algebra to find a
function that satisfies given conditions
Find all zeros of a polynomial function
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Factor Theorem
Example: Determine whetherx+ 4 is a factor of
Yes, it is.
The polynomialx kis a factor of the polynomial
f(k) if and only iff(k) = 0.
( ) = + +4 2
3 48 8 32 f x x x x
4 3 0 48 8 32
123
0
0
4812
0
32
8
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Rational Zeros Theorem
In other words, the numerator is a factor of the constant
term and the denominator is a factor of the first
coefficient.
Ifp/q is a rational number written in lowestterms, and if p/q is a zero off, a polynomial
function with integer coefficients, thenp is a
factor of the constant term and q is a factor of
the leading coefficient.
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Rational Zeros Theorem
Example: For the polynomial function defined by
(a) List all possible rational zeros
For a rational number to be zero,p must be afactor of 4 and q must be a factor of 8:
( ) = + +4 3 2
8 26 27 11 4 f x x x x x
{ } 1, 2, 4p
p
q
1 1 11, 2, 4, , ,2 4 8
pq
{ } , 1, 2, 4, 8q
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Rational Zeros Theorem
Example: For the polynomial function defined by
(b) Find all rational zeros and factorf(x) into linear
factors.
Look at the graph off(x) to judge where it crosses
thex-axis:
( ) = + +4 3 2
8 26 27 11 4 f x x x x x
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Rational Zeros Theorem
Example: For the polynomial function defined by
Use synthetic division to show that 1 is a zero:( ) = + +
4 3 2
8 26 27 11 4 f x x x x x
1 8 26 27 11 4
88 34
347
74
40
( ) ( )( )= + + +3 21 8 34 7 4 f x x x x x
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Rational Zeros Theorem
Example, cont.
Now, we can check the remainder for a zero at 4:
zeros are at 1, 4,
( ) ( )( )= + + +3 2
1 8 34 7 4 f x x x x x
4 8 34 7 4
328 2
81
40
( ) ( )( )( )= + 21 4 8 2 1 f x x x x x
( ) ( )( )( )( )= + + 1 4 4 1 2 1 f x x x x x
1 1
,4 2
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Fundamental Theorem of Algebra
The number of times a zero occurs is referred to as themultiplicity of the zero.
Every function defined by a polynomial of degree
1 or more has at least one complex zero.
A function defined by a polynomial of degree n
has at mostn distinct zeros.
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Fundamental Theorem of Algebra
Example: Find a functionfdefined by a polynomial of
degree 3 that satisfies the following conditions.(a) Zeros of 3, 2, and 5;f(1) = 6
Sincefis of degree 3, there are at most 3 zeros, so thesethree must be it. Therefore,f(x) has the form
( ) ( )( )( )= + +
3 2 5 f x a x x x
( ) ( )( ) ( )( )( )= 3 2 5 f x a x x x
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Fundamental Theorem of Algebra
Example, cont.
We also know thatf(1) = 6, so we can solve for a:
Therefore, or
( ) ( )( )( ) = + +31 1 1 2 51f a
( )( )( )= = 6 2 1 6 12a a
= 12
a
( ) ( )( )( )= + + 1
3 2 52
f x x x x
( ) = + +31 19
152 2
f x x x
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Fundamental Theorem of Algebra
Example: Find a functionfdefined by a polynomial of
degree 3 that satisfies the following conditions.(b) 4 is a zero of multiplicity 3;f(2) = 24
This means that the zero 4 occurs 3 times:
or
( ) ( )( )( )= 4 4 4 f x a x x x
( ) ( )=
3
4 f x a x
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Fundamental Theorem of Algebra
Example, cont.
Since f(2) = 24, we can solve for a:
Therefore, or
( ) ( )= 3
42 2f a
( ) = = 3
24 2 8a a= 3a
( ) ( )= 3
3 4 f x x
( ) = + 3 2
3 36 144 192 f x x x x
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Conjugate Zeros Theorem
This means that if3 + 2iis a zero for a polynomial
function with real coefficients, then it also has 3 2ias a
zero.
Iff(x) defines a polynomial function having only
real coefficients and ifz= a + bi is a zero off(x),where a and b are real numbers, then z= a bi
is also a zero off(x).
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Conjugate Zeros Theorem
Example: Find a polynomial function of least degree
having only real coefficients and zeros 4 and 3 i.The complex number 3 + i must also be a zero, so the
polynomial has at least three zeros and has to be at least
degree 3. We dont know anything else about the
function, so we will leta = 1.
( ) ( )( ) ( )( ) ( )( )= +4 3 3 f x x x i x i( ) ( ) ( ) ( ) ( )( )( )= + + + +24 3 3 3 3 f x x x i x i x i i
( ) ( )( )2 24 3 3 9 f x x x x ix x ix i= + + +
( ) ( )( )= + + = +2 3 24 6 10 2 14 40 f x x x x x x x
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Putting It All Together Example: Find all zeros of
given that 2 + i is a zero.
First, we divide the function by :
( ) = + 4 3 217 55 50 f x x x x x
( )( ) +2x i
+ 2 1 1 17 55 50i
1
2+i
1+i
1+3i
16+3i
3510i
2010i
50
0
( ) ( )( ) ( ) ( ) ( )( )= + + + + + + 3 22 1 16 3 20 10 f x x i x i x i x i
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Putting It All Together Example, cont.
We also know that the conjugate, 2 i is a zero, so we
can divide the remainder by this:
( ) ( )( ) ( ) ( ) ( )( )= + + + + + + 3 2
2 1 16 3 20 10 f x x i x i x i x i
+ + 2 1 1 16 3 20 10i i i i
1
2i
3
63i
10
20+10i
0( ) ( )( ) ( )( )( )= + + 22 2 3 10 f x x i x i x x
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Putting It All Together Example, cont.
Lastly, we can factor or use the quadratic formula to find
our remaining zeros:
So, our zeros are at 2+i, 2i, 5, and 2.
( ) ( )( ) ( )( )( )= + + 22 2 3 10 f x x i x i x x
( )( )+ = + 2 3 10 5 2 x x x x
( ) ( )( ) ( )( )( )( )= + + 2 2 5 2 f x x i x i x x
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Homework Page 337: 5-60 (5s)
HW: 20, 30, 40a, 50, 60