Zeros of Polynomial Functions

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Zeros of Polynomial Functions

Objectives:

1.Use the Fundamental Theorem of Algebra to determine the number of zeros of polynomial functions

2.Find rational zeros of polynomial functions

3.Find conjugate pairs of complex zeros

4.Find zeros of polynomials by factoring

5.Use Descartes’s Rule of Signs and the Upper and Lower Bound Rules to find zeros of polynomials

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In the complex number system, every nth-degree polynomial has precisely “n” zeros.

Fundamental Theorem of Algebra

If f(x) is a polynomial of degree n, where n > 0, then f has at least one zero in the complex number system

Linear Factorization Theorem

If f(x) is a polynomial of degree n, where n > 0, then f has precisely n linear factors

where are complex numbers



f (x)  a_n(x c₁)(x  c₂)...(x c_n)



c₁,c₂,....,c_n

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Zeros of Polynomial Functions

Give the degree of the polynomial, tell how many zeros there are, and find all the zeros



f (x)  x  2

f (x)  x²  6x  9 f (x)  x³  4 x

f (x)  x⁴ 1

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Rational Zero (Root) Test

To use the Rational Zero Test, you should list all rational

numbers whose numerators are factors of the constant term

& whose denominators are factors of the leading coefficient

Once you have all the possible zeros test them using

substitution or synthetic division to see if they work and indeed are a zero of the function (Also, use a graph to help determine zeros to test)

NOTE: It only tests for rational numbers….

factors of the constant term Possible Rational Zeros

factors of leading coefficient



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EXAMPLE: Using the Rational Zero Theorem

List all possible rational zeros of f(x)  15x³  14x²3x – 2.

Solution The constant term is –2 & the leading coefficient is 15.

Factors of the constant term, 2 Possible rational zeros

Factors of the leading coefficient, 15 1, 2

1, 3, 5, 15

 

  

   

Divide 1 and 2

by 1.

Divide 1 and 2

by 3.

Divide 1 and 2

by 5.

Divide 1 and 2 by 15.

There are 16 possible rational zeros. The actual solution set to f(x)  15x³  14x²3x – 2 = 0 is {-1, ^1/³, ²/⁵}, which contains 3 of the 16 possible solutions.

1 2 1 2 1 2

5 5

3 3 15 15

1, 2, , , , , ,

        

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You Try: Using the Rational Zero Theorem

List all possible rational zeros of f(x)  x³ 5x²2x + 8.

Solution The constant term is 8 & the leading coefficient is 1.

Factors of the leading coefficient, 1 1, 2, 4, 8

1

 

   

 

There are 8 possible rational zeros. The actual solution set to f(x)

 x³ 5x²2x + 8 is {-1, 2, 4}, which contains 3 of the 8 possible solutions.

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Roots & Zeros of Polynomials II

Finding the Roots/Zeros of Polynomials (Degree 3 or higher):

• Graph the polynomial to find your first zero/root

• Use synthetic division to find a smaller polynomial

• If the polynomial is not a quadratic follow

the 2 steps above using the smaller polynomial until you get a quadratic.

• Factor or use the quadratic formula to find your remaining zeros/roots

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Example 1:

Find all the zeros of each polynomial function

3 2

10xx x9196First, graph the equation to find the first zero

From looking at the graph you can see that there is a zero at -2

ZERO

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Example 1 Continued

3 2

10xx x9196Second, use the zero you

found from the graph and do synthetic division to find a smaller polynomial

-2 10 9 -19 6

Don’t forget your

remainder should be zero

-20 22 -6 10 -11 3 0

The new, smaller polynomial is: 10 1x x²13

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Example 1 Continued:

10 1x x213 Third, factor or use the quadratic formula to find the remaining zeros.

This quadratic can be factored into:

(5x – 3)(2x – 1)

Therefore, the zeros to the problem are: 3 2

10xx x9196 2, ,3 1

x 5 2

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3 2

( ) 4 21 34

f x  x  x  x 

Using the Rational Zero Theorem: Find all Zeros!

Try listing all possible rational zeros of

Solution The constant term is 34 & the leading coefficient is 1.

Factors of the leading coefficient, 1 1, 2, 17, 34

1

 

   

 

There are 8 possible rational zeros. The actual solution set to f(x)

 x³ 4x²21x34 is {2, 1 + 4i, 14i}, which contains 3 of the 8 possible solutions.

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Rational Zeros

Find the rational zeros.

4 3 2

3 2

( ) 3 6

( ) 8 40 525

( ) 2 3 8 3

f x x x x x

f x x x x

    

   

   

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Find all the real zeros (Hint: start by finding the rational zeros)



f (x)  10x³ 15x²  16x 12

f (x)  3x³ 19x²  33x  9

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Writing a Polynomial given the zeros.

To write a polynomial you must write the zeros out in factored form. Then you multiply the factors together to get your polynomial.

Factored Form: (x – zero)(x – zero). . .

***If it is a polynomial function and a + bi is a root, then a – bi is also a root.

***If it is a polynomial function and is a root, then is also a root

a b a b

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Example 1:

The zeros of a third-degree polynomial are 2 (multiplicity 2) and -5. Write a polynomial.

(x – 2)(x – 2)(x – (-5)) = (x – 2)(x – 2)(x+5)

First, write the zeros in

factored form

Second, multiply the factors out to find your polynomial

( x  2 ) ( xx  2 ) 4 

2

 x  4

( x  5)( x

2

 4 x  4)  x

^{3 2}

 x x  1 6 2 0

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Example 1 Continued

(x – 2)(x – 2)(x+5)

( x  2 ) ( xx  2 ) 4 

2

 x  4

First FOIL or box two of the factors 2

4 4

x x 

X

5

x 3 4x² 4x 5x2 ^²^0x 20

Second, box your answer from above with your remaining

factors to get your polynomial:

3 21620

xx x

ANSWER

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Conjugate Pairs

Complex Zeros Occur in Conjugate Pairs = If a + bi is a zero of the function, the conjugate a – bi is also a zero of the

function

(the polynomial function must have real coefficients) EXAMPLES: Find a polynomial with the given zeros -1, -1, 3i, -3i

2, 4 + i, 4 – i

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So if asked to find a polynomial that has zeros, 2 and 1 – 3i, you would know another root would be 1 + 3i.

Let’s find such a polynomial by putting the roots in factor form and multiplying them together.



^x^²

 

^x^¹^³ⁱ

 

^x^¹^³ⁱMultiply the last two factors together. All i terms should disappear when simplified.



^x^²

 

^x²^^x^³^xi^^x^¹^³ⁱ^³^xi^³ⁱ^⁹ⁱ²



 

^²



²^²^^{1 0}



-1

xxx Now multiply the x – 2 through

20 14

4²

3

xxx

Here is a 3^rd degree polynomial with roots 2, 1 - 3i and 1 + 3i

If x = the root then

(x - the root) is the factor form.



^x^²

 

^x^¹^³ⁱ

 

^x^¹^³ⁱ

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Find ALL the zeros of

Given that 1+3i is a zero!

Example 3:

₄ ₃ ₂

3 6 2 60

x  x  x  

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You Try!

3 2

2 49 98

x  x  x 

Find ALL the zeros of Given that (7i) is a zero!

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STEPS For Finding the Zeros given a Solution

1) Find a polynomial with the given solutions (FOIL) 2) Use long division to divide your polynomial you

found in step 1 with your polynomial from the problem

3) Factor or use the quadratic formula on the answer you found from long division.

4) Write all of your answers out

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Find Roots/Zeros of a Polynomial

If the known root is imaginary, we can use the Complex Conjugates Theorem.

Ex: Find all the roots of f(x)x³5x²7x5 1

If one root is 4 - i.

Because of the Complex Conjugate Thm., we know that another root must be 4 + i.

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Example (con’t)

Ex: Find all the roots of ^f⁽^x⁾^^x³^⁵^x²^⁷^x^^{5 1} If one root is 4 - i.

If one root is 4 - i, then one factor is [x - (4 - i)], and Another root is 4 + i, & another factor is [x - (4 + i)].

Multiply these factors:

[x(4i)][x(4i)](x4i)(x4i)

4 x  i

X

-4

-i

x2 ^^4x ^ix

4x 16 4i

ix _4i i²

2

81 7

 x x

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Example (con’t)

Ex: Find all the roots of ^f⁽^x⁾^^x³^⁵^x²^⁷^x^^{5 1} If one root is 4 - i.

x²8x17

If the product of the two non-real factors is x²8x17

then the third factor (that gives us the real root) is the quotient of P(x) divided by

x²8x17x³5x²7x51 x³5x²7x51

0 x3



The third root

is x = -3

So, all of the zeros are: 4 – i, 4 + i, and -3

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FIND ALL THE ZEROS



f (x)  x⁴  3x³  6x²  2x  60

(Given that 1 + 3i is a zero of f)



f (x)  x³  7x²  x  87

(Given that 5 + 2i is a zero of f)

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More Finding of Zeros



f (x)  x⁵  x³  2x² 12x  8

f (x)  3x³  4 x²  8x  8

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Descartes’s Rule of Signs

Let be a polynomial with real coefficients and

The number of positive real zeros of f is either equal to the number of variations in sign of f(x) or less than that number by an even integer The number of negative real zeros of f is either equal to the number of variations in sign of f(-x) or less than that number by an even integer

Variation in sign = two consecutive coefficients have opposite signs



f (x)  a_nxⁿ  a_n₁x^n¹  ... a₂x²  a₁x  a₀



a₀  0

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EXAMPLE: Using Descartes’ Rule of Signs

Determine the possible number of positive and negative real zeros of f(x)  x³  2x²5x + 4.

Solution

1. To find possibilities for positive real zeros, count the number of sign

changes in the equation for f(x). Because all the terms are positive, there are no variations in sign. Thus, there are no positive real zeros.

2. To find possibilities for negative real zeros, count the number of sign changes in the equation for f(x). We obtain this equation by replacing x with x in the given function.

f(x)  (x)³  2(x)² 5x4

f(x)  x³  2x²5x + 4 This is the given polynomial function.

Replace x with x.

 x³  2x²5x + 4

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EXAMPLE: Using Descartes’ Rule of Signs

Determine the possible number of positive and negative real zeros of f(x)  x³  2x²5x + 4.

Solution

Now count the sign changes.

There are three variations in sign.

# of negative real zeros of f is either equal to 3, or is less than this number by an even integer.

This means that there are either 3 negative real zeros

or 3  2  1 negative real zero.

f(x)  x³  2x²5x + 4

1 2 3

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Descartes’s Rule of Signs

EXAMPLES: describe the possible real zeros



f (x)  3x³  5x²  6x  4 f (x)  3x³  2x²  x  3

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Upper & Lower Bound Rules

Let f(x) be a polynomial with real coefficients and a positive leading coefficient. Suppose f(x) is divided by x – c, using synthetic didvision

If c > 0 and each number in the last row is either positive or zero, c is an upper bound for the real zeros of f

If c < 0 and the numbers in the last row are alternately positive and negative (zero entries count as positive or negative), c is a lower bound for the real zeros of f

EXAMPLE: find the real zeros



f (x)  6x³  4x²  3x  2

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h(x) = x⁴+ 6x³ + 10x² + 6x + 9

1 1 6 10 6 9

1

2 4

6

6 0

0

4 9

Signs are all positive, therefore 1 is an upper bound.

1 of

Factors 9

of

Factors

1

9 3

1



 , ,

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EXAMPLE

You are designing candle-making kits. Each kit contains 25 cubic inches of candle wax and a model for making a pyramid-shaped

candle. You want the height of the candle to be 2 inches less than the length of each side of the candle’s square base. What should the

dimensions of your candle mold be?