Unit 5: Polynomials and Polynomial Functions

Evaluating Polynomial Functions

Objectives: SWBAT identify polynomial functions SWBAT evaluate polynomial functions.

SWBAT find the end behaviors of polynomial functions.

Polynomial ~ multiple monomials in one expression Polynomial Function Notation ~ going from largest degree to

smallest Leading Coefficient ~ number in front of the variable with the

largest degree

Defining Polynomials

Polynomial # of terms Name by # of

terms Degree Name by degree

12 1

Monomial 0 Constant

8𝑥 1

Monomial 1 Linear

4𝑥2 + 3 2

Binomial 2 Quadratic

5𝑥3 + 𝑥2 2

Binomial 3 Cubic

3𝑥2 − 4𝑥 + 6 3

Trinomial 2 Quadratic

3𝑥4 − 4𝑥3 + 6𝑥2 − 7 4

Polynomial 4 Quartic

Decide whether the function is a polynomial function. If so write it in standard form, then state its degree, type, and leading coefficient.

1. 3 2.5 23 4 6f x x x x

Y/N: Standard Form: Degree: Leading Coefficient: # of Terms:

No 3 2.5 23 4 6f x x x x 3 3 4

2. 2 43.7 9f x x x x

Y/N: Standard Form: Degree: Leading Coefficient: # of Terms:

Yes 4 29 3.7f x x x x 4 9 3

3. 3 422 2 3 5x xf x x

Y/N: Standard Form: Degree: Leading Coefficient: # of Terms:

Yes 3 4 22 3 2 5f x x x x 3 3 4

Direct Substitution ~ plug the number directly into each variable and

evaluate

Use direct substitution to evaluate the functions below for the given value.

4. 4 233 2 4; 2f x x x xx 5. 23 2 2 1; 2f x x fx x

4 3 22 3 2 2 2 2 4

2 48 8 8 4

2 52

f

f

f

3 22 2 2 2 2 2 1

2 8 8 4 1

2 11

f

f

f

Polynomial Synthetic Substitution~

1. Write in standard form

2. Write the coefficients in the box, and the x / f(x) term outside the box

3. Add down and multiply outsides

4. Last number is your answer

Use synthetic substitution to evaluate the functions below for the given value.

Perform the following Synthetic substitutions.

1. 4 32 5 4 8 3; f x x x x x 2. 4 3 23 4 2; g x x x x x x

3 2 -5 0 -4 8 -2 1 3 -4 -1 0

(+) 6 3 9 15 (+) -2 -2 12 -22

2 1 3 5 23 1 1 -6 -11 -22

3. 2 45 6 1 4; g x x x x g 4. 4 2 164 8 4

2; f x x x x f

4 5 0 1 6 -1 64 0 -8 -4 0

(+) 20 80 324 1320 32 16 4 0

5 20 81 330 1319 64 32 8 0 0

1

2

Graph Polynomial Functions

Objectives: SWBAT graph polynomial functions SWBAT state the end behavior of polynomial functions

Importance of the leading Coefficient~ tells you all about the end

behavior

Leading Coefficient

Positive Negative

Quadratic

Cubic

Quartic

𝒙 −𝟑 −𝟐 −𝟏 0 𝟏 𝟐 𝟑

𝒇(𝒙) -12 2 4 0 -4 -2 12

Leading Coefficients

Positive Leading

Coefficients

Negative Leading

Coefficients

EVEN DEGREE

Left Side Right Side

x x

What is happening to the graph?

f x f x

Left Side Right Side x x

What is happening to the

graph?

f x f x

ODD DEGREE

Left Side Right Side x x

What is happening to the

graph?

f x f x

Left Side Right Side x x

What is happening to the

graph?

f x f x

Graph the following polynomial functions by creating a table, and then describe

their end behavior.

1. 3 5f x x x

f x as x and f x as x

𝒙 −𝟑 −𝟐 −𝟏 0 𝟏 𝟐 𝟑

𝒇(𝒙) 38 11 0 -1 2 3 -4

𝒙 −𝟑 −𝟐 −𝟏 0 𝟏 𝟐 𝟑

𝒇(𝒙) 38 -2 -2 2 -2 -2 38

𝒙 −𝟑 −𝟐 −𝟏 0 𝟏 𝟐 𝟑

𝒇(𝒙) -142 -29 0 -1 -2 3 -4

2. 3 22 2 1f x x x x

f x as x and f x as x

3. 4 25 2f x x x

f x as x and f x as x

4. 4 3 23 4 1f x x x x x

f x as x and f x as x

Important Parts on the Graph of a Polynomial Function

Objectives: SWBAT identify relative minimum and maximum SWBAT state when a graph is increasing and decreasing

Vocabulary

Relative (Local) Minimum~ a minimum value (where the graph changes direction)

Relative (Local) Maximum~ a maximum value (where the graph changes direction)

Increasing~ Graph’s y value is increasing

(written using it’s x value)

Decreasing~ Graph’s y value is decreasing

(written using it’s x value)

Positive~ when the graph is above the x axis

Negative~ when the graph is below the x axis

1. State where the function is increasing and decreasing.

A. Never Increasing

Decreasing: (−∞, +∞)

B. Increasing: (−8, 0) ∩ (0, + ∞) Decreasing: (−∞, −8)

C. Increasing: (−∞, −8) ∩ (0, 8) Decreasing: (−8, 0)

D. Increasing: (−𝟖, −𝟒) ∩ (𝟎, +∞) Decreasing: (−∞, −𝟖) ∩ (−𝟒, 𝟎)

2. What are the values of the relative maxima and/or minima of the function graphed? Value implies you need the y value)

A. relative maxima: 0

relative minima: −4, 4

B. relative maxima: 𝟏𝟎

relative minima: −𝟏, 𝟐

C. relative maxima: 3.3, 4.7

relative minima: 0

D. relative maxima: 2, 10

relative minima: −1

Sketch a graph of the following polynomial functions by using their local minima or maxima, describe

the end behavior and the intervals in which the function is increasing or decreasing.

3. 3 2( ) 2 3f x x x x 4.

4 3 2( ) 8 15f x x x x

Local Max: Local Max:

Local Min: Local Min:

End Behavior: End Behavior:

f (x) as x and f (x) as x f (x) as x and f (x) as x

Increasing: Positive: Increasing: Positive:

,0.5352 1.8685, 1,0 3, 0,1.7753 4.2248, ,0 0,3 5,

Decreasing: Negative: Decreasing: Negative:

0.5352,1.8685 , 1 0,3 ,0 1.7753,4.2248 3,5

Add, Subtract, and Multiply Polynomials

Objectives: SWBAT Add, Subtract, and Multiply Polynomials

Adding polynomials vertically and horizontally.

Add like terms

1. 3 2 3 2(3x 2x 4x 6) (x 5x 3) 2.

3 2 2(2y 7y 6y) ( 4y 3y 9)

3 3

3

2 2

2

3

4

2

4

5

7

6 3

3

4x

x

x

x

xx

x

x

3

23

23

27y

4

6y

3y

2y

y

0

9

9 3

0

3y y

y

2y

Subtracting polynomials vertically and horizontally.

Subtract like terms

3. 3 2 3 2(7x 6x 4x 7) (6x 3x 7x 5) 4.

2 2(8y 5y 11) (12y 9y 3)

3 3

3

2 2

2

6 3 77 57

39

6

2

4x x x xx

xx

x

x

2

2

2

8y

12y

5y

9y

11

3

4y 4 4 1y

Multiplying polynomials vertically and horizontally.

Multiply the terms

5. 2(3x x 4)(x 2) 6.

2(x 3)(x 2x 5)

3 2

2

2

2

3

2

2

3 4 6 2 8

( ) 4

4

2

3

2

5

3

3 3

2

2

(

2

)

8

4

x

x x

x

x x x

x

x

x x

x

x x x

x x x

3 2

2

23

2

( 2 5)

x ( 3

15

1

3

52

6

1 51

)

x

x

x

x

x

x

x

x

x

x

x

Simplify the following expressions using the method of your choice.

7. 2 3 2(8t 6 5t ) (2t 3t 7) 8.

2 2(4p 6p 6) (8p 7p 4) 9. 2(x 2)(6x 3x 5)

2

23

235t 3t(8 6 7

1

2t

8t

(

2t

) )t

2t

2 2

2

4p 8( 6p 7p

13

6 4) ( )

p

p

212p

2

3 2 2

3 2

( 2)(6 3 5)

6 3 5 12 6 10

6 9 11 10

x x x

x x x x x

x x x

10. (x 3)(x 7)(x 1)

3 2 2

3 2

2

(x 1)

(

x

(x 3)

x

(x

4x 4x 21x 21

x 5x 17x 2

7

x 1

)

)x

1

x 4 21

SPECIAL PRODUCT PATTERNS

****Shortcuts if you can remember them*****

Sum and Difference:

2 2(a b)(a b) a b Example: 2(x 2)(x x2) 4

a = x and b = 2

Square of a Binomial:

2 2 2(a b) a 2ab b Example: 2 2y y) 16(y 4 8

a = y and b = 4

2 2 2(a b) a 2ab b Example: 2 22 9(3 2p 42 p p) 1

a = 3p and b = 2

Cube of a Binomial:

3 3 2 2 3(a b) a 3a b 3ab b Example: 3 3 2x(x 1) 3x 3x 1

a = x and b = 1

3 3 2 2 3(a b) a 3a b 3ab b Example: 3 3 2r 9(r r 7r 273) 2

a = r and b = 3

Perform the following Polynomial Multiplication. If a rule can be use, stay which one, label your “a” and “b,” and then use it.

11. 2(6q 3r) 12.

3(2m 5) 13. (x 3)(x 6)(x 5)

2 2

2 2

a 2ab b

a 6q b 3r

36q 36qr 9r

2 2

2

a 2ab b

a 2m b 5

4m 20m 25

2

3 2 2

3 2

No Rule

PEMDAS(Left to Right)

(x 3)(x 6)(x 5)

x 3x 18 x 5

x 5x 3x 15x 18x 90

x 2x 33x 90

Factoring and Solving Polynomial Equations

Objectives: SWBAT Factor polynomials

Prime polynomial - Factored completely -

Can’t be factoring any further each parenthesis can’t be factored Any further, check for perfect

Squares / cubes

SPEICAL FACTORING PATTERNS

Sum of two cubes:

3 3 2 2a b (a b)(a ab b ) Example: 3x 8 3 3 22 2 2 4x x x x

a = x and b = 2

Difference of two cubes:

3 3 2 2a b (a b)(a ab b ) Example: 38x 1 3 3 22 1 2 1 4 2 1x x x x

a = 2x and b =1

Factor the sum or difference of cubes.

2 terms only so factor by Perfect Square or Perfect Cube

1. 3z 125 2. 481y 192y 3. 38x 64

3 3

2

5

5 5 25

z

z z z

4

3

3 3

2

81 192

3

3 27 64

3 4

3 3 4 9 12 16

y y

y

y y

y

y y y y

3

3

3 3

2

8 64

8

8 8

2

8 2 2 4

x

x

x

x x x

Factor by grouping.

4 terms only so factor by Grouping

4. 3 2x 2x 9x 18 5. 3 2x 2x 25x 50

3 2

2

2

2

2

2 9 18

9

2 9 2

9 2

9

3 3 2

x x x

x

x x x

x x

x can be factored

x x x

3 2

2

2

2

2

2 25 50

5

2 5 2

5 2

5

5 5 2

x x x

x

x x x

x x

x can be factored

x x x

Factor polynomials in quadratic form.

3 terms so factor by using U substitution

Look for GCF U Substitution

6. 4 214 45x x 7. 7 5 33y 15y 18y

2

2

2 2

14 45

5 9

5 9

5 5 3 3

u x

u u

u u

x x

y y y y

7 5 3

3

4 2

2

2

2 2

3

3y 15y 18y

3y

y 5y 6

u y

u 5u 6

u 3 u 2

y 3 y 2

3y y 3 y 3 y 2 y 2

Z.A.R.S.~

Zeros, answers, roots, and solutions Factor, set each parenthesis equal to zero, and solve.

Solve a polynomial equation.

8. 4 2x 9 10x 9. 5 32x 24x 14x

4 2

2

2

2 2

10 9 0

3

10 9 0

9 1 0

9 1 0

3 3 1 1 0

3 0 1 0

3 0 1 0

1, 1,3, 3

x x

terms U Substitution

u x

u u

u u

x x

x x x x

x x

x x

x

5 3

4 2

2

2

2 2

2 14 240

2

2 7 12 0

3

7 12 0

4 3 0

4 3 0

2 2 3 3 0

2 0 3 0

2 0 3 0

2, 2, 3, 3

x x x

x

x x x

terms U Substitution

u x

u u

u u

x x

x x x x

x x

x x

x

Long and Synthetic Division

Objectives: SWBAT Factor and Divide Polynomials

Review:

1. 12 502 2. 1 5 45 7. .

1

22

1

12 502

10

1041

12

2

4

48

Answer

Remaind

er

:

3

45

15 457

0

7

15

0

3

0

0

7

Answer

7 Remainder

:

Polynomial Long Division~

1. Works just like long division 2. Write the polynomial in standard from including zeros for missing

terms 3. Multiply the divisor 4. Subtract the two expressions together 5. Bring down the next turn and repeat until the highest exponent

meets the exponent of the quotient term

Divide the following expressions using polynomial long division.

3. 3 2x 6x 9 x 4

Put into Standard form first include missing parts

See what is needed to make the first terms Divider and quotient term the same

Multiply the x2 to each term of the divisor and subtract

Just like Long division, bring down the next term, and find out how many times the leading term of the divisor goes into the new expression

Continue the same process until you get a term where the leading term of the divisor can no

longer go into the new expression. That will final expression will be a remainder

2

3 2

2

3 2

2

2

2

2 0

2 8

8

8 9

8

6 0 9

23

23

3

4

4

2

82

4

x

x

x x

Answ

x

x

x x

x x

x

x

x

x

er x

x

x

4. Divide 4 33 5 4 6f x x x x by 2 3 5x x

Put into Standard form first include missing parts

See what is needed to make the first terms

Divider and quotient term the same

Multiply the 3x2 to each term of the divisor and subtract

Just like Long division, bring down the next term, and find out how many times the leading term of the divisor goes into the new expression

Continue the same process until you get a term where the leading term of the divisor can no

longer go into the new expression. That will final expression will be a remainder

Answer will be 2

2

25 93 4 3

3 5

xx x

x x

4

2

2

4 3 2

2

3 2

3

3

3

3 5 3 5 0 4 6

9 15

4 15

-

x

x

x

x x x x x

x x

x x

2

2

4

3 2

2

2

3

4 3

3 2

2

3

3

3 5 3 5

4

4

4 12 20

3

9

4 6

5

5

0

1

4 1

x

x

x x x

x x

x

x

x x

x x x x

x

x

x

16x

3

2

2

4 3

3

4

2

2

3

2

23 5 3 5 0

4

4

4 1

4 6

3

3 9 15

4 1

3

2

5

20

3

x

x

x x

x

x

x x

x x x

x

x x

x

x x

2

2 61

25 9

3

6

9 15

x x

x x

x

2 4 3 23 5 3 5 0 4 6x x x x x x

5. Divide 4 2 24 17 9 18 2 2x x x x

Synthetic Division

Short cut division that can be used when your divide a term like x – a or x + a

Exactly like Synthetic substitution only you switch a sign in the divisor

Each term before will the last will be a coefficient starting with a power one

less than the original expression

The last term will be the remainder

Divide the following expressions using synthetic division.

1. Divide 3 25 7 2f x x x x by 2x 2. 3 22 8 5 3x x x x

2 1 5 -7 2 -3 2 1 -8 5

(+) 2 14 14 (+) -6 15 -21

1 7 7 16 2 -5 7 -16

2

3

2 4 3 2

4

3

3 2

2

2

132

2

2 0 2 4 0 17 9

0

0 13 9

0 0

18

4 0 4

13 9 1

0

8

x

x x x

x x

x

x x x x x x

x x x

x x

2

2

2

13 0 13

13 9 312

2 2

9

2

31

x x

xAns x

x

x

wer

2 167 7

2Answer x x

x

2 16

2 5 73

Answer x xx

3. 4 25 2 15 10 2x x x x

-2 5 0 2 -15 10

-10 20 -44 118

5 -10 22 -59 128

4. 4 3 23 5 2 3 2 3 2x x x x x

Because it isn’t just an x, it’s a 3x, we can’t use Synthetic division, so we go back to long division

3 2 1285 10 22 59

2Answer x x x

x

2

3

4 3 2

4

2

2

3 2

3

2

3

3 2 3 5 2 3

0

2

0 3

0 0

3

2

2

2

1

3

3

x

x

x

x x x x

x

x

x

x x

x

x x

x

x x

3 2

3 2

0

x

x

3 2 1Answer x x

Apply the Remainder and Factor Theorems

Objectives: SWBAT Factor a polynomial given a factor

SWBAT Find all the zeroes of a polynomial given a zero

FACTOR THEOREM or Throw Me a Bone Theorem

Given a factor, use synthetic division

Then find the other factors by factoring

Given polynomial f(x) and a factor of f(x), factor f(x) completely.

1. 3 22 11 3 36 3 f x x x x x; 2. 3 23 8 3 2 2f x x x x x ;

3 2 –11 3 36 –2 3 8 3 –2 6 –15 –36 –6 –4 2

2 –5 –12 0 3 2 –1 0

22 5 12x x

23 2 1x x

2

2

2 12

2 12

2 4 3 4

2 3

8 3

2 3

4

8 3

4 3

x

x

x x x

x

x x

x x

x x x

x

2

2

2

3 1

3 1

3 1

3 2

1 1

3 1

3 1

3

2

1

3 1

1

1

1

x

x

x x x

x x

x

x x x

x x

x

x

x

3. 3 26 5 12 4f x x x x x ;

4 1 –6 5 12

4 –8 –12

1 –2 –3 0

2 2 3x x

2 2 3

3 1 4

x x

x x x

Finding Zeros~ Factor and set each parenthesis equal to zero If it can’t factor, use quadratic formula

Given polynomial function f(x) and a zero of f(x), find the other zeros.

4. 3 28 48 2f x x x x ; 5. 3 22 3 13 12 3f x x x x ;

–2 1 0 –28 –48 3 2 –3 –13 12

–2 4 48 6 9 –12

1 –2 –24 0 2 3 -4 0

2 2 24x x

22 3 4x x

6 4 0

6 0 4 0

6, 4, 2

x x

x x

x x x

2

2

4

2

3 3 4 2 4

2 2

3 9 32

4

3 41

4

3 41 3 41, , 3

4 4

b b ac

a

x x x

6. 3 22 23 60 3f x x x x x ;

3 1 –2 –23 60

3 3 –60

1 1 -20 0

2 20x x

5 4 0

5 0 4 0

5, 4, 3

x x

x x

x x x

Apply the Rational Zero Theorem

Objectives: SWBAT find the zeros of polynomials (with an a=1) without being given a factor

Vocabulary Review

Leading Coefficient~ number in front of the variable with the

highest degree

Constant~ number without a variable

RATIONAL ZERO THEOREM

If n

n 1 0f (x) a x ... a x a has integer coefficients, then every rational zero of f has the

following form: p each factor of the constant term

q each factor of the leading coefficient

List the possible rational zeros 𝒇 using the rational zero theorem.

1. 3 22 11 12f x x x x 2. 4 3 24 3 9 10f x x x x x

Factors of the constant term: Factors of the constant term:

1,2,3,4,6,12 1,2,5,10

Factors of the leading coefficient: Factors of the leading coefficient:

1 1,2,4

Possible rational zeros: Possible rational zeros:

1 2 3 4 6 12, , , , ,

1 1 1 1 1 1

1 2 5 10 1 2 5 10 1 2 5 10, , , , , , , , , , ,

1 1 1 1 2 2 2 2 4 4 4 4

Simplified list of zeros: Simplified list of zeros:

1, 2, 3, 4, 6, 12 1 5 1 5

1, 2, 4, 10, , , , ,2 2 4 4

3. 3 22 3 11 6f x x x x

Factors of the constant term: 1,2,3,6

Factors of the leading coefficient: 1,2

Possible rational zeros: 1 2 3 6 1 2 3 6

, , , , , , ,1 1 1 1 2 2 2 2

Simplified list of zeros: 1 3

1, 2, 3, 6, ,2 2

STEPS TO FINDING ALL REAL ZEROS (Finding the Bone) when 𝒂 = 𝟏

1. Arrange the polynomial in standard form,

2. Write down all the factors of the constant term. These are all the possible values of p,

3. Write down all the factors of the leading coefficient. These are all the possible values

of q,

4. Write down all the possible values of . Remember that since factors can be

negative, and - must both be included. Simplify each value and cross out any duplicates.

5. Use synthetic division to determine the values of for which P( ) = 0 . These are all the

rational roots of the function

****HINT!!! You may have to do this more than once****

Find all real zeros of the function.

4. 3 28 11 20f x x x x 5. 4 215 10 24f x x x x

Simplified list of zeros: Simplified list of zeros: 1, 2, 4, 5, 10, 20 1, 2, 4, 5, 6, 8, 12, 24

–1 1 –8 11 20 –3 1 0 –15 –10 24

–1 9 –20 –3 9 18 –24

1 –9 20 0 1 –3 –6 8 0

2 9 20 0x x

–2 1 –3 –6 8

–2 10 –8

1 –5 4 0

5 4 0

5 0 4 0

5, 4, 1

x x

x x

x x x

2 5 4 0x x

1 4 0

1 0 4 0

1, 4, 2, 3

x x

x x

x x x x

6. 3 24 15 18f x x x x

Simplified list of zeros:

1, 2, 3, 6, 9, 18

1 1 –4 –15 18 1 –3 –18

1 –3 –18 0

2 3 18 0x x

6 3 0

6 0 3 0

6, 3, 1

x x

x x

x x x

Find Rational Zeros

Objectives: SWBAT find the zeros of polynomials without being given a factor

STEPS TO FINDING ALL REAL ZEROS WHEN THE LEADING

COEFFICIENT ≠ 𝟏

1. Arrange the polynomial in standard form,

2. Write down all the factors of the constant term. These are all the possible values of p,

3. Write down all the factors of the leading coefficient. These are all the possible values

of q,

4. Write down all the possible values of . Remember that since factors can be

negative, and - must both be included. Simplify each value and cross out any duplicates.

5. Use synthetic division to determine the values of for which P( ) = 0 . These are all the

rational roots of the function

****HINT!!! You may have to do this more than once****

Find all real zeros of the following function.

1. 3 24 12 15 h x x x x

Simplified list of zeros:

15 15

1, 2, 3, 5, 15, ,2 4

–1 4 –12 –1 15

–4 16 –15

4 –16 15

24 16 15 0x x

2

2

4 15

4 15

2 2 5 3 2 5

2 3 2 5

2

10 6

10 6

3

3 0 2 5

5

0

, , 12 2

x

x

x x x

x x

x

x x

x x

x x x

x

2. 4 3 29 12 26 11 6 f x x x x x

Simplified list of zeros:

1 2 2

1, 2, 3, 6, , ,3 3 9

2

3 9 12 –26 –11 6

–6 –4 20 –6

9 6 –30 9 0

1

3 9 6 –30 9

3 3 –9

9 9 –27 0

2

2

9 9 27 0

3 0

x x

x x

3. 4 3 210 11 42 7 12 f x x x x x

Simplified list of zeros:

12 6 4 3 2 1 1

1, 2, 3, 6, 12, , , , , , ,5 5 5 5 5 5 10

1

2 10 –11 –42 7 12

–5 8 17 –12

10 –16 –34 24 0

3

5 10 –16 –34 24

6 –6 –24

10 –10 –40 0

2

2

4

2

1 1 4 1 3

2 1

1 1 12

2

1 13

2

1 13 1 13 2 1, , ,

2 2 3 3

b b ac

a

x x x x

2

2

4

2

1 1 4 1 4

2 1

1 1 16

2

1 17

2

1 17 1 17 1 3, , ,

2 2 2 5

b b ac

a

x x x x

2

2

10 10 40 0

4 0

x x

x x

4. Cassie is building a wooden square sandbox for a local playground. She wants the volume of the box to be 16 cubic feet. She decides that the height of the box should be x

feet, and that she would like the length of each side of the square base to be three feet longer than the height. What dimensions should she build her sandbox?

2

3 2

3 3 16

6 9 16

6 9 16 0

x x x

x x x

x x x

Simplified list of zeros: 1, 2, 4, 8, 16

1 1 6 9 –16

1 7 16

1 7 16 0

2 7 16 0x x

2

2

4

2

7 7 4 1 16

2 1

7 49 64

2

7 15

2

7 15

2

7 151,

2

1

b b ac

a

i

ix x

Only Works

Apply the Fundamental Theorem of Algebra

Objectives: SWBAT find the zeros of polynomials

Repeated Solutions~ when a factor comes up twice or more, it is a repeated solution

FUNDAMENTAL THEOREM OF ALGEBRA

Theorem: If f(x) is a polynomial of degree n where n greater than 0, then the

equation f(x) = 0 has at least one solution in the set of complex numbers.

Corollary: If f(x) is a polynomial of degree n, then the equation f(x) = 0 has exactly

n solutions provided each solution that is repeated twice is counted as twice solutions, each solution repeated three times is counted as three solutions and so on.

Find the number of solutions or zeroes for each equation or function.

Highest degree states the number of solutions you should have.

1. 23 5x 4 0 0x x 2 2. 4 3 2f (x) 8x 18x 2x 7

3 4

Find all the zeroes for each function below.

3. 5 4 3 2g(x) x 4x 4x 10x 13x 14

Simplified list of zeros:

1, 2, 7, 14

–1 1 –4 4 10 –13 –14

–1 5 –9 –1 14

1 –5 9 1 –14 0

–1 1 –5 9 1 –14

–1 6 –15 14

1 –6 15 –14 0

2 1 –6 15 –14

2 –8 14

1 –4 7 0

2

2

2

4 7 0

4

2

4 4 4 1 7

2 1

4 16 28

2

4 12

2

4 2 3

2

2 3

1, 1, 2, 2 3, 2 3

x x

b b ac

a

i

i

x i i

4. 5 4 2f (x) x 2x 8x 13x 6

Simplified list of zeros:

1, 2, 3, 6

–2 1 –2 0 8 –13 6

–2 8 –16 16 –6

1 –4 8 –8 3 0

1 1 –4 8 –8 3

1 –3 5 –3

1 –3 5 –3 0

1 1 –3 5 –3

1 –2 3

1 –2 3 0

2

2

2

2 3 0

4

2

2 2 4 1 3

2 1

2 4 12

2

2 8

2

2 2 2

2

2 2

1, 1, 2, 2 2, 2 2

x x

b b ac

a

i

i

x i i

BEHAVIOR NEAR ZEROS

Use your calculator to approximate the real zeroes of the function below.

5. 6 5 4 3 2h(x) x 2x 3x 10x 6x 8x 8

Use your calculator, then hit second

Calculate to find the zeros on each side.

Scenario

Positive

Real Zeros (from the

positive sign

change)

Negative

Real Zeros (From the

negative sign

change)

Imaginary

Zeros (Total

Zeros – positive

zeroes – negative

zeroes)

Total

Zeros (highest

degree)

1 3 3 0 6

2 3 1 2 6

3 1 3 2 6

4 1 1 4 6

Apply Descartes’ Rule of Signs

Objectives: SWBAT apply Descartes’ Rule of Signs for Solving Polynomials

DESCARTES’ RULE OF SIGNS

For any polynomial function with real coefficients…

The number of positive real zeroes of f is equal to the number of sign changes in the coefficients of f(x), or less than this by an even number.

The number of negative real zeroes of f is equal to the number of sign changes in the coefficients of f(-x), or less than this by an even number.

Determine the possible numbers of positive real zeros, negative real zeros, and imaginary zeros.

1. 6 45 3 22x 10x 6x 8f ( xx) x 3 x 8

Positive Sign Changes

3 Sign changes Negative Sign Changes

2

6 5

5 3

3

6

4 2

42 3 1

( ) 2 3 10

0 8

6 8 8

6( ) 8

f x x x x x x x

x x xf xxx x

3 sign changes

Scenario Positive

Real Zeros (from the

positive sign

change)

Negative

Real Zeros (From the

negative sign

change)

Imaginary

Zeros (Total

Zeros – positive

zeroes – negative

zeroes)

Total

Zeros (highest

degree)

1 4 0 0 4

2 2 0 2 4

3 0 0 4 4

2. 34 22 3) 1( 68x xg x x x

Positive Sign Changes

4 Sign changes

Negative Sign Changes

4 3 2

4 3 2

( ) 2 8 6 3 1

( ) 2 8 6 3 1

g x x x x x

g x xx x x

0 sign changes

COMPLEX CONJUGATES THEOREM If f is a polynomial function with real coefficients, and a +bi is an imaginary zero of f, then a -bi is also a zero of f.

IRRATIONAL CONJUGATES THEOREM

Similar to above, if a b is an imaginary zero of f, then a b is also a

zero of f, when a and b are rational numbers.

Write a polynomial function of least degree that has rational coefficients, a leading

coefficient of one, and the given zeros.

3. –1, 2, 4 4. 2 and 3i 5. 1, –2, 3

2

2

2 2

4 2 3 2

4 3 2

1 2 3 3

1 0 2 0 3 0 3 0

1 2 3 3

1 2 3 3 9

1 2 3

2 3

3 3 2 6

3 6

x x x x

x x x x

x x x x

x x x x x

x x x

x x x

x x x x x

x x x x

2 2

2

3 2 2

2 3 3

2 0 3 0 3 0

2 3 3

2 3 3 9

2 9

9 2 16

x x i x i

x x i x i

x x i x i

x x ix ix i

x x

x x x

2

3 2 2

3 2

1 2 4

1 0 2 0 4 0

1 2 4

2 4

4 4 2 8

5 6 8

x x x

x x x

x x x

x x x

x x x x x

x x x

Analyze Graphs of Polynomial Functions

Objectives: SWBAT find the zeros of polynomials without being given a factor

Relative (Local) Maximum~ Relative (Local) Minimum~

a maximum value (where the a minimum value (where

graph changes direction) graph changes direction)

Use a graphing calculator to graph the polynomial function. Identify the end

behavior, the x-intercepts, and the points where the local maximums and local minimums occur for the graph.

1. 5 3 2f (x) x 4x x 2

End Behavior: f (x) as x and f (x) as x

x-intercepts: See diagram

Local Minimums: See diagram

Local Maximums: See diagram

Increasing: , 1.627 1.458, Domain: ,

Decreasing: 1.627,0 0,1.458 Range: ,

Positive: 2.159,1.747 Negative: , 2.159 1.747,