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 epression Polynomial Function Notation ~ going from largest degree to smallest Leading Coefficient ~ number in front of the variable with the largest degree Defining Polynomials Polynomial 1 8 4 + 5 + 4 + 6 4 4 + 6 7 # of terms 1 1 4 Name by # of terms Degree Name by degree Monomial 0 Constant Monomial 1 Linear Binomial Quadratic Binomial Cubic Trinomial Quadratic 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..5 f 4 6 Y/N: Standard Form: Degree: Leading Coefficient: # of Terms: No.5 f 4 6 4
. 4 f.7 9 Y/N: Standard Form: Degree: Leading Coefficient: # of Terms: Yes 4 f 9.7 4 9 4. f 5 Y/N: Standard Form: Degree: Leading Coefficient: # of Terms: Yes 4 f 5 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. f 4; 5. f 1; f f f 48 8 8 4 f 5 4 4 f f 8 8 4 1 f 11 1 Polynomial Synthetic Substitution~ 1. Write in standard form. Write the coefficients in the bo, and the / f() term outside the bo. 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. 4 4 1. f 5 4 8;. g 4 ; -5 0-4 8-1 -4-1 0 (+) 6 9 15 (+) - - 1-1 5 1 1-6 -11-4 4 1. g 5 6 1; g4 4. 64 8 4 ; f f 4 5 0 1 6-1 1 64 0-8 -4 0 (+) 0 80 4 10 16 4 0 5 0 81 0 119 64 8 0 0
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
Leading Coefficients Positive Leading Negative Leading Coefficients Coefficients EVEN DEGREE Left Side Right Side What is happening to the graph? f f Left Side Right Side What is happening to the graph? f f ODD DEGREE Left Side Right Side Left Side Right Side What is happening to the graph? What is happening to the graph? f f f f Graph the following polynomial functions by creating a table, and then describe their end behavior. 1. f 5 1 0 1 f() -1 4 0-4 - 1 f as and f as
. f 1 1 0 1 f() 8 11 0-1 -4 f as and f as. 4 f 5 1 0 1 f() 8 - - - - 8 f as and f as 4. 4 f 4 1 1 0 1 f() -14-9 0-1 - -4 f as and f as
Important Parts on the Graph of a Polynomial Function Vocabulary Objectives: SWBAT identify relative minimum and maimum SWBAT state when a graph is increasing and decreasing Relative (Local) Minimum~ a minimum value (where the graph changes direction) Relative (Local) Maimum~ a maimum value (where the graph changes direction) Increasing~ Graph s y value is increasing (written using it s value) Decreasing~ Graph s y value is decreasing (written using it s value) Positive~ when the graph is above the ais Negative~ when the graph is below the ais 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)
. What are the values of the relative maima and/or minima of the function graphed? Value implies you need the y value) A. relative maima: 0 relative minima: 4, 4 B. relative maima: 10 relative minima: 1, C. relative maima:., 4.7 relative minima: 0 D. relative maima:, 10 relative minima: 1 Sketch a graph of the following polynomial functions by using their local minima or maima, describe the end behavior and the intervals in which the function is increasing or decreasing.. f ( ) 4. f ( ) 8 15 4 Local Ma: Local Ma: Local Min: Local Min: End Behavior: End Behavior: f() as and f() as f() as and f() as Increasing: Positive: Increasing: Positive:,0.55 1.8685, 1,0, 0,1.775 4.48,,0 0, 5, Decreasing: Negative: Decreasing: Negative: 0.55,1.8685, 1 0,,0 1.775, 4.48,5
Add, Subtract, and Multiply Polynomials Objectives: SWBAT Add, Subtract, and Multiply Polynomials Adding polynomials vertically and horizontally. Add like terms 1. ( 4 6) ( 5 ). (y 7y 6y) ( 4y y 9) 4 5 7 4 6 4 y 0y y 7y 4y y 6y 0 y 9 y 9 Subtracting polynomials vertically and horizontally. Subtract like terms. (7 6 4 7) (6 7 5) 4. (8y 5y 11) (1y 9y ) 7 6 6 477 5 9 8y 1y 5y 11 9y 4y 4y 14 Multiplying polynomials vertically and horizontally. Multiply the terms 5. ( 4)( ) 6. ( )( 5) ( ) ( 4 4 4 4 6 8 5 ) 8 ( 5) ( ) 6 15 5 11 15
Simplify the following epressions using the method of your choice. 7. (8t 6 5t ) (t t 7) 8. (4p 6p 6) (8p 7p 4) 9. ( )(6 5) (8t 65t ( t t 7 t ) ) t 8t 1 ( 4p 6p 8p 7p 1p 6) ( 4) 1p ( )(6 5) 6 5 1 6 10 6 9 11 10 10. ( )( 7)( 1) ( ) ( 7) ( 1) 4 1 5 17 ( 1) 4 4 1 1 1
SPECIAL PRODUCT PATTERNS ****Shortcuts if you can remember them***** Sum and Difference: (a b)(a b) a b Eample: ( )( a = and b = ) 4 Square of a Binomial: (a b) a ab b Eample: (y 4 8 ) y y16 a = y and b = 4 (a b) a ab b Eample: ( ) 9 1 p p p 4 a = p and b = Cube of a Binomial: (a b) a a b ab b Eample: ( 1) 1 a = and b = 1 (a b) a a b ab b Eample: (r ) r 9r 7r 7 a = r and b = Perform the following Polynomial Multiplication. If a rule can be use, stay which one, label your a and b, and then use it. 11. (6q r) 1. (m 5) 1. ( )( 6)( 5) a ab b a 6q b r 6q 6qr 9r a ab b a m b 5 4m 0m 5 No Rule PEMDAS(Left to Right) ( )( 6)( 5) 18 5 5 15 18 90 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
Sum of two cubes: SPEICAL FACTORING PATTERNS a b (a b)(a ab b ) Eample: 4 8 a = and b = Difference of two cubes: a b (a b)(a ab b ) Eample: 8 1 1 14 1 a = and b =1 Factor the sum or difference of cubes. 1. terms only so factor by Perfect Square or Perfect Cube z 15. 4 81y 19y. 8 64 z 5 z 5z 5z 5 4 81y 19 y y 7 y 64 y 4 y y 4 9y 1y 16 y 8 64 8 8 8 8 4 Factor by grouping. 4. 4 terms only so factor by Grouping 9 18 5. 5 50 9 18 9 9 9 9 can be factored 5 50 5 5 5 5 can be factored 5 5
Factor polynomials in quadratic form. terms so factor by using U substitution 6. Look for GCF 14 45 7. 4 U Substitution 7 5 y 15y 18y u u u5u9 5 9 y 5 y 5 y y 14u45 y 15y 18y y 7 5 4 y 5y 6 u y u u y y u 5u 6 y y y y y Z.A.R.S.~ Zeros, answers, roots, and solutions Factor, set each parenthesis equal to zero, and solve. Solve a polynomial equation. 8. 4 9 10 9. 5 4 14 4 10 9 0 terms U Substitution u u u u 10u 9 0 9 1 0 9 1 0 1 1 0 0 1 0 0 1 0 1, 1,, 5 14 4 0 4 7 1 0 terms U Substitution u u u u 7u1 0 4 0 4 0 0 0 0 0 0,,,
Long and Synthetic Division Objectives: SWBAT Factor and Divide Polynomials Review: 1. 1 50. 1. 5 45. 7 41 1 50 48 1 10 Remainder 10 Answer : 41 1 0 15 457 45 07 0 7 Remainder 7 Answer : 0 15 Polynomial Long Division~ 1. Works just like long division. Write the polynomial in standard from including zeros for missing terms. Multiply the divisor 4. Subtract the two epressions together 5. Bring down the net turn and repeat until the highest eponent meets the eponent of the quotient term
Divide the following epressions using polynomial long division.. 6 9 4 4 8 6 0 9 4 0 8 Answer 8 9 8 8 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 to each term of the divisor and subtract Just like Long division, bring down the net term, and find out how many times the leading term of the divisor goes into the new epression Continue the same process until you get a term where the leading term of the divisor can no longer go into the new epression. That will final epression will be a remainder
4. Divide 4 f 5 4 6 by 5 Put into Standard form first include missing parts 4 5 5 0 4 6 See what is needed to make the first terms Divider and quotient term the same 4 5 5 0 4 6-9 15 4 4 15 Multiply the to each term of the divisor and subtract 5 5 4 0 4 9 15 4 4 1 5 4 1 0 4 4 1 4 6 4 16 4 5 5 0 4 9 15 5 4 1 4 6 4 0 Just like Long division, bring down the net term, and find out how many times the leading term of the divisor goes into the new epression Continue the same process until you get a term where the leading term of the divisor can no longer go into the new epression. That will final epression will be a remainder 16 6 915 5 9 Answer will be 4 5 9 5
5. Divide 4 4 17 9 18 1 0 18 4 0 4 0 17 9 4 4 0 4 0 1 9 0 0 0 1 91 8 1 01 9 1 Answer 1 9 1 Synthetic Division Short cut division that can be used when your divide a term like a or + a Eactly 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 epression The last term will be the remainder Divide the following epressions using synthetic division. 1. Divide f 5 7 by. 8 5 1 5-7 - 1-8 5 (+) 14 14 (+) -6 15-1 1 7 7 16-5 7-16 16 Answer 7 7 16 Answer 5 7
. 5 4 15 10-5 0-15 10-10 0-44 118 5-10 -59 18 18 Answer 5 10 59 4. 4 5 Because it isn t just an, it s a, we can t use Synthetic division, so we go back to long division 4 5 Answer 4 1 0 0 0 0 1 0
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() and a factor of f(), factor f() completely. 1. f 11 6 ;. f 8 ; 11 6 8 6 15 6 6 4 5 1 0 1 0 5 1 1 8 1 81 4 4 4 4 11 1 1 1 1 1 1 1 1 1 1
. f 6 5 1; 4 4 1 6 5 1 4 8 1 1 0 1 4 Finding Zeros~ Factor and set each parenthesis equal to zero If it can t factor, use quadratic formula
Given polynomial function f() and a zero of f(), find the other zeros. 4. f 8 48 ; 5. f 1 1; 1 0 8 48 1 1 4 48 6 9 1 1 4 0-4 0 4 4 6 4 0 6 0 4 0 6, 4, b b 4ac a 4 4 9 4 41 4 41 41,, 4 4
6. f 60; 1 60 60 1 1-0 0 0 5 4 0 5 0 4 0 5, 4,
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 f() a... a a has integer coefficients, then every rational zero of f has the n n 1 0 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. 4 1. f 11 1. f 4 9 10 Factors of the constant term: Factors of the constant term: 1,,,4,6,1 1,,5,10 Factors of the leading coefficient: Factors of the leading coefficient: 1 1,,4 Possible rational zeros: 1 4 6 1,,,,, 1 1 1 1 1 1 Simplified list of zeros: 1,,, 4, 6, 1 Possible rational zeros: 1 5 10 1 5 10 1 5 10,,,,,,,,,,, 1 1 1 1 4 4 4 4 Simplified list of zeros: 1 5 1 5 1,, 4, 10,,,,, 4 4
. f 11 6 Factors of the constant term: 1,,,6 Factors of the leading coefficient: 1, Possible rational zeros: Simplified list of zeros: 1 6 1 6,,,,,,, 1 1 1 1 1 1,,, 6,,
STEPS TO FINDING ALL REAL ZEROS (Finding the Bone) when a = 1 1. Arrange the polynomial in standard form,. Write down all the factors of the constant term. These are all the possible values of p,. 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 4. f 8 11 0 5. f 15 10 4 Simplified list of zeros: Simplified list of zeros: 1,, 4, 5, 10, 0 1,, 4, 5, 6, 8, 1, 4 1 1 8 11 0 1 0 15 10 4 1 9 0 9 18 4 1 9 0 0 1 6 8 0 9 0 0 1 6 8 10 8 1 5 4 0 5 4 0 5 0 4 0 5, 4, 1 5 4 0 1 4 0 1 0 4 0 1, 4,,
6. f 4 15 18 Simplified list of zeros: 1,,, 6, 9, 18 1 1 4 15 18 1 18 1 18 0 18 0 6 0 6 0 0 6,, 1
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,. Write down all the factors of the constant term. These are all the possible values of p,. 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. h 4 1 15 1. Simplified list of zeros: 15 15 1,,, 5, 15,, 4 1 4 1 1 15 4 16 15 4 16 15 4 1615 0 4 10615 4 10615 5 5 5 0 5 0 5,, 1
f 9 1 6 11 6. 4 Simplified list of zeros: 1 1,,, 6,,, 9 9 1 6 11 6 1 6 4 0 6 9 6 0 9 0 9 6 0 9 9 9 9 7 0 9 9 7 0 0 b b 4ac a 1 1 1 4 1 1 11 1 1 1 1 1 1 1,,, f 10 11 4 7 1. 4 Simplified list of zeros: 1 6 4 1 1 1,,, 6, 1,,,,,,, 5 5 5 5 5 5 10 1 10 11 4 7 1 5 5 8 17 1 10 16 4 4 0 10 16 4 4 6 6 4 10 10 40 0 b b 4ac a 1 1 1 4 1 4 1 116 1 17 1 17 1 17 1,,, 5 10 10 40 0 4 0
4. Cassie is building a wooden square sandbo for a local playground. She wants the volume of the bo to be 16 cubic feet. She decides that the height of the bo should be 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 sandbo? 16 6 9 16 6 9 16 0 Simplified list of zeros: 1,, 4, 8, 16 1 1 6 9 16 1 7 16 1 7 16 0 716 0 b b 4ac a 1 7 7 4 1 16 7 49 64 7 15 7 i 15 7 i 15 1, Only 1 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() is a polynomial of degree n where n greater than 0, then the equation f() = 0 has at least one solution in the set of comple numbers. Corollary: If f() is a polynomial of degree n, then the equation f() = 0 has eactly 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. 5 4 0 0. 4 f () 8 18 7 4
Find all the zeroes for each function below.. 5 4 g() 4 4 10 1 14 Simplified list of zeros: 1,, 7, 14 1 1 4 4 10 1 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 1 6 15 14 8 14 1 4 7 0 4 7 0 b b 4ac a 4 4 417 1 4 16 8 4 1 4 i i 1, 1,, i, i
4. 5 4 f () 8 1 6 Simplified list of zeros: 1,,, 6 1 0 8 1 6 8 16 16 6 1 4 8 8 0 1 1 4 8 8 1 5 1 5 0 1 1 5 1 1 0 0 b b 4ac a 41 1 4 1 8 i i 1, 1,, i, i
BEHAVIOR NEAR ZEROS Use your calculator to approimate the real zeroes of the function below. 5. 6 5 4 h() 10 6 8 8 Use your calculator, then hit second Calculate to find the zeros on each side.
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(), 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(-), or less than this by an even number. Determine the possible numbers of positive real zeros, negative real zeros, and imaginary zeros. 1. f( 6 5 4 ) 10 6 88 Positive Sign Changes Sign changes Negative Sign Changes 6 5 4 f ( ) 10 6 8 8 f 6 5 4 ( ) 10 6 8 8 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 0 6 1 6 1 6 4 1 1 4 6
. g ( 6 4 ) 8 1 Positive Sign Changes 4 Sign changes Negative Sign Changes 4 g( ) 8 6 1 g 4 ( ) 8 6 1 0 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 0 4 0 0 4 4 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 bis an imaginary zero of f, then a bis 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.. 1,, 4 4. and i 5. 1,, 1 1 4 1 0 0 4 0 1 4 4 4 4 8 5 6 8 i i 0 i 0 i 0 i i i i 9i 9 9 16 1 0 0 0 0 1 1 9 1 4 6 4 6
Analyze Graphs of Polynomial Functions Objectives: SWBAT find the zeros of polynomials without being given a factor Relative (Local) Maimum~ a maimum value (where the graph changes direction) Relative (Local) Minimum~ a minimum value (where graph changes direction) Use a graphing calculator to graph the polynomial function. Identify the end behavior, the -intercepts, and the points where the local maimums and local minimums occur for the graph. 1. 5 f () 4 End Behavior: f() as and f() as -intercepts: See diagram Local Minimums: Local Maimums: See diagram See diagram Increasing:, 1.67 1.458, Domain:, Decreasing: 1.67,0 0,1.458 Range:, Positive:.159,1.747 Negative:,.159 1.747,