Obj. 6 Zeros of Polynomial Functions (Presentation)

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


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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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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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(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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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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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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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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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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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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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

Obj. 6 Zeros of Polynomial Functions (Presentation)

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