Enhanced Instructional Transition Guide

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1 1-1 Enhanced Instructional Transition Guide High School Courses Unit Number: 7 /Mathematics Suggested Duration: 9 days Unit 7: Polynomial Functions and Applications (15 days) Possible Lesson 1 (6 days) Possible Lesson (9 days) Possible Lesson (9 days) This lesson is one approach to teaching the State Standards associated with this unit. Districts are encouraged to customize this lesson by supplementing with districtapproved resources, materials, and activities to best meet the needs of learners. The duration for this lesson is only a recommendation, and districts may modify the time frame to meet students needs. To better understand how your district is implementing CSCOPE lessons, please contact your child s teacher. (For your convenience, please find linked TEA Commissioner s List of State Board of Education Approved Instructional Resources and Midcycle State Adopted Instructional Materials.) Lesson Synopsis: Students investigate the algebraic behavior of polynomial functions. Through a series of activities using graphing calculators, students explore the relationships between a polynomial function, its factors, and the graph s x-intercepts. Students learn algebraic methods to find the zeros and factors of a function, such as synthetic substitution, synthetic division, and long division of polynomials. Students are introduced to several important rules regarding polynomials, such as the Factor Theorem and the Fundamental Theorem of Algebra. TEKS: The Texas Essential Knowledge and Skills (TEKS) listed below are the standards adopted by the State Board of Education, which are required by Texas law. Any standard that has a strike-through (e.g. sample phrase) indicates that portion of the standard is taught in a previous or subsequent unit. The TEKS are available on the Texas Education Agency website at P.1 /Knowledge and Skills. The student defines functions, describes characteristics of functions, and translates among verbal, numerical, graphical, and symbolic representations of functions, including polynomial, rational, power (including radical), exponential, logarithmic, trigonometric, and piecewise-defined functions. The student is expected to: P.1D Recognize and use connections among significant values of a function (zeros, maximum values, minimum values, etc.), points on the graph of a function, and the symbolic representation of a function. P. /Knowledge and Skills. The student interprets the meaning of the symbolic representations of functions and operations on functions to solve meaningful problems. The student is expected to: page 1 of 88

2 1-1 Enhanced Instructional Transition Guide High School Courses Unit Number: 7 /Mathematics Suggested Duration: 9 days P.B Perform operations including composition on functions, find inverses, and describe these procedures and results verbally, numerically, symbolically, and graphically. Performance Indicator(s): High School Mathematics Unit 7 PI Formulate polynomial functions that would meet the certain criteria such as the following: Write a cubic polynomial function to match the graph shown below. Write a quadratic polynomial equation with roots of + 5i and 5i. Standard(s): P.1A, P.1D, P.1E, P.A ELPS ELPS.c.1E High School Mathematics Unit 7 PI Use graphical and algebraic methods to determine the exact values of all zeros, roots, and factors of polynomial functions such as the following: Describe the relationship between the graph, degree, zeros, and factors of a polynomial function. Recognize the nature of the roots (complex and real) of a polynomial function from the zeros of its graph. Make connections between zeros, roots, and factors. Display all results using a graphic organizer. Standard(s): P.1A, P.1D, P.B ELPS ELPS.c.1C page of 88

3 1-1 Enhanced Instructional Transition Guide High School Courses Unit Number: 7 /Mathematics Suggested Duration: 9 days Key Understanding(s): Polynomial functions can be determined using various characteristics of the function, such as zeros and roots. The Fundamental Theorem of Algebra states that the degree of a polynomial equation must be the same as the number of complex zeros, although these solutions are not always distinct x intercepts on the function s graph. These solutions must be determined algebraically and can be irrational and/or imaginary. Misconception(s): Some students may think that a number can be classified as only real, rational, or complex and not realize the categories overlap, such as 8 is rational, real, and complex. Underdeveloped Concept(s): Some students may have difficulty simplifying complex expressions that arise when the quadratic formula yields negative radicands. Some students may have difficulty (and make errors) when multiplying trinomials or expanding expressions that involve complex and imaginary numbers. Vocabulary of Instruction: complex number complex roots cubic function degree of a function depressed polynomial double root factor imaginary number irrational number polynomial function polynomial long division quadratic function quartic function rational number rational root theorem real number real roots root (or solution) synthetic substitution x-intercept zero page of 88

4 1-1 Enhanced Instructional Transition Guide High School Courses Unit Number: 7 /Mathematics Suggested Duration: 9 days Materials List: graphing calculator (1 per student) graphing calculator with display (1 per teacher) scissors (1 per student) Attachments: All attachments associated with this lesson are referenced in the body of the lesson. Due to considerations for grading or student assessment, attachments that are connected with Performance Indicators or serve as answer keys are available in the district site and are not accessible on the public website. Quotation Solving KEY Quotation Solving Synthetic Substitution KEY Synthetic Substitution Function Factors KEY Function Factors Graphs and Factors KEY Graphs and Factors Sports Puns KEY Sports Puns Finding a Way KEY page of 88

5 1-1 Enhanced Instructional Transition Guide High School Courses Unit Number: 7 /Mathematics Suggested Duration: 9 days Finding a Way Polynomial Equations KEY Polynomial Equations Solution Scramble KEY Solution Scramble The Missing Factor KEY The Missing Factor The Complex Nature of Things KEY The Complex Nature of Things Exact Zeros KEY Exact Zeros Equation Solver KEY Equation Solver Do-It-Yourself Equations Polynomial Algebra KEY Polynomial Algebra PI Suggested Day Suggested Instructional Procedures Notes for Teacher page 5 of 88

6 1-1 Enhanced Instructional Transition Guide High School Courses Unit Number: 7 /Mathematics Suggested Duration: 9 days 1 Topics: Polynomial equations Engage 1 Students match solutions to polynomial equations. ATTACHMENTS Teacher Resource: Quotation Solving KEY (1 per teacher) Handout: Quotation Solving (1 per student) Instructional Procedures: 1. Distribute handout: Quotation Solving to each student. Place students in pairs. Instruct students to work with their partner to complete the handout. Allow students about 1 15 minutes to complete the handout, and monitor to observe methods used by students to determine solutions. Remind students that more than one solution may work for an equation. All but one of these equations has multiple solutions.. Facilitate a class discussion of student results, emphasizing methods used to determine the solutions. Possible answers may include: MATERIALS graphing calculator (1 per student) TEACHER NOTE For handout: Quotation Solving instruct students to show work by substituting in all values to check solutions. Substitution Evaluate the function at the given values of x, checking each with trial and error. Table Type the related function in the calculator s Y= menu to generate a table of values. Graph Graph the function to find its zeros. Topics: Synthetic substitution Explore/Explain 1 ATTACHMENTS Teacher Resource: Synthetic Substitution KEY (1 per teacher) Teacher Resource: Synthetic page 6 of 88

7 1-1 Enhanced Instructional Transition Guide High School Courses Unit Number: 7 /Mathematics Suggested Duration: 9 days Suggested Day Suggested Instructional Procedures Students evaluate solutions to polynomial equations using synthetic substitution. Instructional Procedures: 1. Emphasize to students that none of the methods in Engage 1 were very fast. Each required a fair amount of time to enter the many terms of the polynomial into a calculator, and search for the correct solution. In these cases, an alternate method can be used that is often much faster that using a calculator. It s called synthetic substitution.. Distribute handout: Synthetic Substitution to each student. Refer students to the top of page 1. Display teacher resource: Synthetic Substitution, and facilitate a class discussion of synthetic substitution, modeling using the Example problem.. Instruct students to work with their partner to complete problems 1 6 on handout: Synthetic Substitution. Allow students time to complete the problems. Monitor and assess students to check for understanding. Using teacher resource: Synthetic Substitution, facilitate a class discussion of student results, clarifying any misconceptions.. Instruct students to work independently to complete problems 7 1 on handout: Synthetic Substitution. This may be completed as homework, if necessary. Notes for Teacher Substitution (1 per teacher) Handout: Synthetic Substitution (1 per student) MATERIALS graphing calculator (1 per student) TEACHER NOTE Why does synthetic substitution work? The process is derived from a method of repeatedly factoring x from the terms of a polynomial. For example: Note, to evaluate the polynomial in this factored form, you would start with (the leading coefficient), then multiply by x, add -17, multiply by x again, then add - 59, multiply by x, and add. These are the same steps involved in using synthetic substitution. - Topics: Polynomial functions ATTACHMENTS Teacher Resource: Function Factors page 7 of 88

8 1-1 Enhanced Instructional Transition Guide High School Courses Unit Number: 7 /Mathematics Suggested Duration: 9 days Suggested Day Suggested Instructional Procedures Polynomial equations Explore/Explain Students investigate connections between graphs and zeros of polynomial functions and factors and solutions of polynomial equations. Instructional Procedures: 1. Place students in pairs. Distribute handout: Function Factors to each student. Instruct students to complete problems 1. Display teacher resource: Function Factors, and facilitate a class discussion of student results. Ask: Is the first graph quadratic, cubic, or quartic? (quadratic) How do you know? Answers may vary. You can tell by the shape of its graph (a parabola). Also, if you multiplied the two factors together, you d get an x term; etc. Where are the function s x-intercepts? (at x = - and ) How do you know? Answers may vary. You could determine zeros of each factor or you could determine the intercepts along the x-axis; etc. Is the second graph quadratic, cubic, or quartic? (cubic) How do you know? Answers may vary. You can tell by the shape of its graph. It has two points of inflection; etc. Where are the function s x-intercepts? (at x = -,, and ) How do you know? Answers may vary. You could determine zeros of each factor or you could determine the intercepts along the x-axis; etc. Notes for Teacher KEY (1 per teacher) Teacher Resource: Function Factors (1 per teacher) Handout: Function Factors (1 per student) Teacher Resource: Graphs and Factors KEY (1 per teacher) Teacher Resource: Graphs and Factors (1 per teacher) Handout: Graphs and Factors (1 per student) Teacher Resource: Sports Puns KEY (1 per teacher) Handout: Sports Puns (1 per student) MATERIALS graphing calculator (1 per student) graphing calculator with display (1 per teacher) TEACHER NOTE Students should arrive at the rule that, if x r is a factor of a polynomial function, then x = r is an x- page 8 of 88

9 1-1 Enhanced Instructional Transition Guide High School Courses Unit Number: 7 /Mathematics Suggested Duration: 9 days Suggested Day Suggested Instructional Procedures. Instruct students to work with their partner to complete the remaining problems on handout: Function Factors. Allow students time to complete the problems. Monitor and assess students to check for understanding. Using teacher resource: Function Factors, facilitate a class discussion of student results, clarifying any misconceptions. Emphasize relationships between factors and graphs of polynomial functions and exceptions to the normal pattern. Notes for Teacher intercept. Why is this true? It is because of the Zero Product Rule (to be discussed later in the lesson). This rule says that if you have a factored expression equal to zero, then solutions arise by setting each factor equal to zero. For example: What is the relationship between the factors and degree? (The number of factors is the same as the degree, as long as squared factors are counted twice.) What is the relationship between the factors and the x-intercepts? Answers may vary. The numbers are opposites; etc. Give an example of a factor and the corresponding x-intercept. Answers may vary. On problem 5, the factor is x plus, and the x-intercept is at negative. The other factor is x minus 5, and the x-intercept is at positive 5; etc. This method works whether there are three, four, or more factors.. Distribute handout: Graphs and Factors to each student. Refer students to the top of page 1. Display teacher resource: Graphs and Factors, and facilitate a class discussion of the Factor Theorem, modeling the Example problem. Instruct students to work with their partner to complete Parts A and B problems 1 6. Part A has students determine the factors and part B has students check their answers by multiplying the factors back together and comparing to the original function. Allow students time to complete the problems. Monitor and assess students to check for understanding. Using teacher resource: Graphs and Factors, facilitate a class discussion of student results, clarifying any misconceptions. Emphasize problems 6 as students may make some common omissions. page 9 of 88

10 1-1 Enhanced Instructional Transition Guide High School Courses Unit Number: 7 /Mathematics Suggested Duration: 9 days Suggested Day Suggested Instructional Procedures. Distribute handout: Sports Puns to each student. Instruct students to work independently to complete the handout. This may be completed as homework, if necessary. Notes for Teacher Topics: Polynomial functions Characteristics of polynomial functions Explore/Explain Students determine effects of parameters on graphical representations of polynomial functions. Instructional Procedures: 1. Place students in pairs. Distribute handout: Finding a Way to each student. Instruct students to complete problem 1. Display teacher resource: Finding a Way, and facilitate a class discussion of student results and the effects of a on the graph of a function. As seen in the previous activity, when trying to determine the rule or equation for a polynomial function, sometimes it is not enough to know only the x-intercepts. There are other graphical features to consider. For example, the graph may have been stretched, compressed, or reflected. These attributes are affected by a constant multiplier (a), which is in fact the leading coefficient.. Instruct students to work with their partner to complete problems 6 on handout: Finding a Way. Allow students time to complete the problems. Monitor and assess students to check for understanding. Using teacher resource: Finding a Way, facilitate a class discussion of student results, clarifying any misconceptions. ATTACHMENTS Teacher Resource: Finding a Way KEY (1 per teacher) Teacher Resource: Finding a Way (1 per teacher) Handout: Finding a Way (1 per student) MATERIALS graphing calculator (1 per student) graphing calculator with display (1 per teacher) TEACHER NOTE Problems 6 9 on handout: Finding a Way ask students to find the polynomial function to match a graph based on its x-intercepts. In certain cases such as 6 & 7, regression may also be used to determine a function rule from the coordinates of page 1 of 88

11 1-1 Enhanced Instructional Transition Guide High School Courses Unit Number: 7 /Mathematics Suggested Duration: 9 days Suggested Day Suggested Instructional Procedures. Refer students to the top of page. Using teacher resource: Finding a Way, facilitate a class discussion of determining a polynomial function given x-intercepts and a point that is passed through. Model the Example problem. Instruct students to work with their partner to complete problems 7 1. Allow students time to complete the problems. Monitor and assess students to check for understanding. Using teacher resource: Finding a Way, facilitate a class discussion of student results, clarifying any misconceptions.. Instruct students to work independently to complete problem 11 on handout: Finding a Way. This may be completed as homework, if necessary. Notes for Teacher points. However, when a double root is present as on 8 & 9, regression does not work, since not enough points are listed. 5 Topics: Polynomial equations Explore/Explain Students use factors of polynomial equations to determine solutions. Instructional Procedures: 1. Place students in pairs. Distribute handout: Polynomial Equations to each student. Refer students to the top of page 1. Display teacher resource: Polynomial Equations, and facilitate a class discussion of the Zero Product Rule, modeling the Example problem.. Instruct students to work with their partner to complete problems 1 6 on handout: Polynomial Equations. Allow students time to complete the problems. Monitor and assess students to check for understanding. Using teacher resource: Polynomial ATTACHMENTS Teacher Resource: Polynomial Equations KEY (1 per teacher) Teacher Resource: Polynomial Equations (1 per teacher) Handout: Polynomial Equations (1 per student) Teacher Resource: Solution Scramble KEY (1 per teacher) Handout: Solution Scramble (1 per student) MATERIALS page 11 of 88

12 1-1 Enhanced Instructional Transition Guide High School Courses Unit Number: 7 /Mathematics Suggested Duration: 9 days Suggested Day Suggested Instructional Procedures Equations, facilitate a class discussion of student results, clarifying any misconceptions. On many of the equations, one factor cannot be solved using simple addition and/or division. In these cases, students must use square roots and the quadratic formula. If necessary, review these equation-solving skills. (See Teacher Note.). Refer students to the top of page. Using teacher resource: Polynomial Equations, facilitate a class discussion of other methods to solve polynomial equations. Instruct students to work with their partner to complete problems 7 1. Allow students time to complete the problems, monitoring to check for student understanding. Using teacher resource: Polynomial Equations, facilitate a class discussion of student results, clarifying any misconceptions. Note that the factors on problems 9 & 1 are somewhat ambiguous. Students may not to know the answers to these problems immediately. The main purpose for these problems in this activity is to remind students that besides rational roots, others are possible (e.g., irrational and complex). The methods for finding these types of missing factors will be addressed in the next activity.. Distribute handout: Solution Scramble to each student. Instruct students to work independently to complete the handout. This may be completed as homework, if necessary. Notes for Teacher graphing calculator (1 per student) graphing calculator with display (1 per teacher) TEACHER NOTE Remind students that in order to solve quadratic equations when there is no first degree x-term, you can isolate the x and then take the square root of both sides: When an x term and an x-term are present, however, the quadratic formula can apply. For example, in x + x 7 =, a = 1, b =, and c = -7. page 1 of 88

13 1-1 Enhanced Instructional Transition Guide High School Courses Unit Number: 7 /Mathematics Suggested Duration: 9 days Suggested Day Suggested Instructional Procedures Notes for Teacher 6 Topics: ATTACHMENTS Polynomial equations Long division of polynomial expressions Synthetic substitution Explore/Explain 5 Students use long division and synthetic substitution to determine factors of polynomial equations. Teacher Resource: The Missing Factor KEY (1 per teacher) Teacher Resource: The Missing Factor (1 per teacher) Handout: The Missing Factor (1 per student) Instructional Procedures: 1. Facilitate a class discussion of problem 1 from handout: Polynomial Equations. Ask: What is the degree of the polynomial in this equation? ( rd degree or cubic) If we graph the related function, how many x-intercepts does it have? Where? (one x-intercept at x = 1.75) What factor do we know? (x 1.75) Does this factor need to be squared? Is it a double root? (no) MATERIALS graphing calculator (1 per student) graphing calculator with display (1 per teacher) TEACHER NOTE In polynomial long division, when subtracting the polynomials, it is often easier to consider adding the page 1 of 88

14 1-1 Enhanced Instructional Transition Guide High School Courses Unit Number: 7 /Mathematics Suggested Duration: 9 days Suggested Day Suggested Instructional Procedures There must be another factor. What is known about the other factor? (It must have an x in it.). Place students in pairs. Distribute handout: The Missing Factor to each student. Instruct students to solve the arithmetic division problem at the top of the page. Allow students time to complete the problem. Display teacher resource: The Missing Factor, and facilitate a class discussion of polynomial long division by making comparisons to steps students used in their division problem and modeling the example problem.. Instruct students to work with their partner to complete problems 1 6 on handout: The Missing Factor. Allow students time to complete the problems. Monitor and assess students to check for understanding. Using teacher resource: The Missing Factor, facilitate a class discussion of student results, clarifying any misconceptions.. Refer students to the top of page on handout: The Missing Factor. Using teacher resource: The Missing Factor, facilitate a class discussion to analyze questions 7 9. Ask: Notes for Teacher opposite, or changing all the signs of the terms in the polynomial being subtracted. For example, to complete x + x (x 5x) Rewrite the problem as x + x + x + 5x TEACHER NOTE The process of synthetic substitution can often be used in place of polynomial long division, but not always. To use synthetic methods, the factor must be of the form x r. If a factor has an x- coefficient other than 1, long division must be used. How are these items different from the previous problems? (No factor is provided.) How can one of the factors be found? (By finding the x-intercepts, and relating these to the factors.) 5. Refer students to the top of page on handout: The Missing Factor. Using teacher resource: The Missing Factor, facilitate a class discussion of the connections between long division and synthetic substitution. Point out that, in these circumstances, synthetic page 1 of 88

15 1-1 Enhanced Instructional Transition Guide High School Courses Unit Number: 7 /Mathematics Suggested Duration: 9 days Suggested Day Suggested Instructional Procedures substitution provides the coefficients of the missing factor, sometimes called the depressed polynomial. 6. Instruct students to work independently to complete problems 1 1 on handout: The Missing Factor. This may be completed as homework, if necessary. Notes for Teacher 7-8 Topics: Polynomial equations Explore/Explain 6 Students develop a process for determining factors and solutions to polynomial equations. Instructional Procedures: 1. Place students in pairs. Distribute handout: The Complex Nature of Things to each student.. Instruct students to work independently to complete problems 1, and when complete, check their work with their partner. Display teacher resource: The Complex Nature of Things, and facilitate a class discussion of student results. Ask: How would you describe the solutions to problem 1? (A double root of 1 and a rational root of -8/.) How are the solutions to problem different? (Two irrational roots (square roots) and a rational root of -5.) How are the solutions to problem different? (Two imaginary solutions (negative ATTACHMENTS Teacher Resource: The Complex Nature of Things KEY (1 per teacher) Handout: The Complex Nature of Things (1 per student) Teacher Resource: Exact Zeros KEY (1 per teacher) Teacher Resource: Exact Zeros (1 per teacher) Handout: Exact Zeros (1 per student) Teacher Resource: Equation Solver KEY (1 per teacher) Handout: Equation Solver (1 per student) MATERIALS graphing calculator (1 per student) graphing calculator with display (1 per page 15 of 88

16 1-1 Enhanced Instructional Transition Guide High School Courses Unit Number: 7 /Mathematics Suggested Duration: 9 days Suggested Day Suggested Instructional Procedures numbers under the radical) and a rational root of.) Which of these solutions are real numbers and why? (Problems 1 and, because they do not have solutions that contain imaginaries.) Why are the numbers that contain i not considered real numbers? Answers may vary. The complex number system is made up of real and non-real numbers. In a real number, the imaginary part zeros out. The non-real numbers contain a non-zero imaginary part. These are non-real (because of the imaginary part), but still complex; etc. Which of the real solutions are called rational? Only the solutions that are whole numbers, integers, decimals, or fractions. They must be able to be written as a numerator over a denominator. What do we call the real solutions with square roots? (irrational) Which type of solutions show up as x-intercepts? (only the real numbers) Notes for Teacher teacher) TEACHER NOTE For handout: Exact Zeros, students are given a polynomial with one of its zeros, and asked to find exact values of the other zeros. The process is similar to the previous activity: A) Verify that the value is a zero using synthetic substitution, B) Use the results to write the polynomial s factors, and C) find the other zeros by solving the equation formed by setting the depressed polynomial equal to zero. This often involves using the quadratic formula.. Refer students to page of handout: The Complex Nature of Things. Using teacher resource: The Complex Nature of Things, facilitate a class discussion of the Fundamental Theorem of Algebra and the nature of roots, modeling problems 5 9. TEACHER NOTE You can use different number-sorting activities to help students classify the nature of numbers. page 16 of 88

17 1-1 Enhanced Instructional Transition Guide High School Courses Unit Number: 7 /Mathematics Suggested Duration: 9 days Suggested Day Suggested Instructional Procedures Instruct students to work with their partner to complete problems 1 1. Allow students time to complete the problems. Monitor and assess students to check for understanding. Using teacher resource: The Complex Nature of Things, facilitate a class discussion of student results. Ask students a series of questions such as the following for graph 1 on page : Notes for Teacher What is the degree of this polynomial? (Degree is 5.) How many zeros should it have? (It should have 5.) How many show up as x-intercepts on the graph? (three) Do any count twice? Are any of the double roots? (no) How many zeros are left? (Two are left.) What must be the nature of these zeros? (They must be non-real.) TEACHER NOTE To help students consider all the possible combinations of real, double, and non-real zeros, imagine the different scenarios that would result from translating a function up or down.. Instruct students to work with their partner to complete problems 15 on handout: The Complex Nature of Things. Allow students time to complete the problems, monitoring to check for student understanding. Using teacher resource: The Complex Nature of Things, facilitate a class discussion of student results, clarifying any misconceptions. A B One real zero, and two nonreal zeros One double root, and one page 17 of 88

18 1-1 Enhanced Instructional Transition Guide High School Courses Unit Number: 7 /Mathematics Suggested Duration: 9 days Suggested Day Suggested Instructional Procedures Notes for Teacher 5. Distribute handout: Exact Zeros to each student. Refer students to the top of page 1. Display teacher resource: Exact Zeros, and facilitate a class discussion of determining all zeros of a polynomial function, and modeling the Example problem on pages Instruct students to work with their partner to complete problems 1 5 on handout: Exact Zeros. Allow students time to complete the problems. Monitor and assess students to check for understanding. Using teacher resource: Exact Zeros, facilitate a class discussion of student results, clarifying any misconceptions. 7. Distribute handout: Equation Solver to each student. Instruct students to work independently to complete the handout. This may be completed as homework, if necessary. C other real zero Three distinct (different) real zeros Topics: ATTACHMENTS Polynomial equations Elaborate 1 Students create and verify a fourth-degree polynomial equation with rational coefficients. Instructional Procedures: 1. Distribute handout: Do-It-Yourself Equations and a pair of scissors to each student.. Instruct students to follow the instructions on the page to create polynomial equations from solutions selected by the individual student. Handout: Do-It-Yourself Equations (1 per student) MATERIALS graphing calculator (1 per student) graphing calculator with display (1 per teacher) scissors (1 per student). After students have developed their original equations from the solutions they selected, page 18 of 88

19 1-1 Enhanced Instructional Transition Guide High School Courses Unit Number: 7 /Mathematics Suggested Duration: 9 days Suggested Day Suggested Instructional Procedures instruct students to cut off the equation at the bottom of the page.. Place students in pairs. Instruct students to exchange equations with their partner and individually solve each other s equations. After solving the equation, instruct students to return the equation back to their partner and check each other s solutions. Notes for Teacher 9 Evaluate 1 Instructional Procedures: 1. Assess student understanding of related concepts and processes by using the Performance Indicator(s) aligned to this lesson. Performance Indicator(s): ATTACHMENTS Teacher Resource (optional): Polynomial Algebra KEY (1 per teacher) Handout (optional): Polynomial Algebra PI (1 per student) page 19 of 88

20 1-1 Enhanced Instructional Transition Guide High School Courses Unit Number: 7 /Mathematics Suggested Duration: 9 days Suggested Day Suggested Instructional Procedures Notes for Teacher High School Mathematics Unit 7 PI Formulate polynomial functions that would meet the certain criteria such as the following: Write a cubic polynomial function to match the graph shown below. MATERIALS graphing calculator (1 per student) TEACHER NOTE As an optional assessment tool, use handout (optional): Polynomial Algebra PI. Write a quadratic polynomial equation with roots of + 5i and 5i. Standard(s): P.1A, P.1D, P.1E, P.A ELPS ELPS.c.1E page of 88

21 1-1 Enhanced Instructional Transition Guide High School Courses Unit Number: 7 /Mathematics Suggested Duration: 9 days Suggested Day Suggested Instructional Procedures Notes for Teacher High School Mathematics Unit 7 PI Use graphical and algebraic methods to determine the exact values of all zeros, roots, and factors of polynomial functions such as the following: Describe the relationship between the graph, degree, zeros, and factors of a polynomial function. Recognize the nature of the roots (complex and real) of a polynomial function from the zeros of its graph. Make connections between zeros, roots, and factors. Display all results using a graphic organizer. Standard(s): P.1A, P.1D, P.B ELPS ELPS.c.1C /6/1 page 1 of 88

22 Quotation Solving KEY Unit: 7 Lesson: For each, circle the x-values that solve the given equations. Cross out the values that are not solutions. The codes under the correct values can be used to reveal a mathematical quote. Equation 1) x 6x + 1x 8 ) x x 1x+ ) x + x 1x + = ) x 1x + 9 Possible Solutions x = x = x = 1 ( = IS) ( = ON) ( = CH) x = x = x = - ( = DE) ( = RE) ( = ST) 1 x = 5 x = x = -5 ( = TI) ( = AT) ( = A) x = - x = x = 1 ( = TH) ( = AT) ( = MA) 5) x + 6x 9x 1 = 6) x 17x 59x+ = x = -7 x = -1 x = ( = IS) (Ω = CS) ( = OT) 1 x = x = x = 8 ( = IN) ( = G) ( = E) 7) x x 16x+ 1= x = x = 1.5 x = - ( = SP) ( = D) ( = UP).7 % of all ST AT IS TI CS A RE Ω MA DE UP ON TH E SP OT. 1, TESCCC 1/1/1 page 1 of 1

23 Quotation Solving Unit: 7 Lesson: For each, circle the x-values that solve the given equations. Cross out the values that are not solutions. The codes under the correct values can be used to reveal a mathematical quote. Equation 1) x 6x + 1x 8 ) x x 1x+ ) x + x 1x + = ) x 1x + 9 5) x + 6x 9x 1 6) x 17x 59x+ = Possible Solutions x = x = x = 1 ( = IS) ( = ON) ( = CH) x = x = x = - ( = DE) ( = RE) ( = ST) 1 x = 5 x = x = -5 ( = TI) ( = AT) ( = A) x = - x = x = 1 ( = TH) ( = AT) ( = MA) x = -7 x = -1 x = ( = IS) (Ω = CS) ( = OT) 1 x = x = x = 8 ( = IN) ( = G) ( = E) 7) x x 16x+ 1= x = x = 1.5 x = - ( = SP) ( = D) ( = UP).7 % of all Ω. Steven Wright 1, TESCCC 1/1/1 page 1 of 1

24 Synthetic Substitution KEY Unit: 5 Lesson: 1 The method described below can be used to quickly evaluate a polynomial function at given values of x without the aid of technology. Example: If = x + 6x 9x 1, find f (). x-value coefficients First, write the coefficients (in order, according to descending powers) Write the x-value to be substituted up and to the left Bring down (or, copy) the leading coefficient below. Then repeat these two steps: Multiply by the x-value, and write the answer under the next coefficient to the right. Add down, and write the answer below. Continue multiplying when you move over, and adding when you move down. The final value will be the function evaluated at the given x-value. Or, in this case, f () =. Using the same function, evaluate f () and f ( 7). f() =, and f(-7) = Use synthetic substitution to evaluate each function at the given values of x. 1) = x 6x + 1x 8 ) = x 6x + 1x 8 ) g ( x) = x + x 1x + Find f (). f () = -5 Find f (). f () = Find g (). g () = ) h ( x) = x 17x 59x+ 5) h ( x) = x 17x 59x+ 6) g ( x) = x + x 1x + Find h ( ). h (-) = 5 Find h (8). h (8) = Find g ( 1 ). g ( 1 ) = ½ , TESCCC 1/1/1 page 1 of

25 Synthetic Substitution KEY Unit: 5 Lesson: 1 7) = x + 5x 1x x+ 8) = x + 5x 1x x+ Find f (). f () = 9 Find f ( 6). f (-6) = For m ( x) = x 7, find m(). To use synthetic substitution when terms of a certain power are not present, a coefficient of zero must be used to fill the place For example, the coefficients of x 7 would be,, and -7 (because the polynomial has no x term). 1 1 Continue using synthetic substitution to evaluate each polynomial. 9) = x 1x+ 16 1) a ( x) = x + x + Find f ( ). f (-) = Find a (). a () = ) ( ) 1 7 f x = x x + x 1) y ( x) = x 17x + 16 Find f (). f () = 8 Find y ( ). y (-) = , TESCCC 1/1/1 page of

26 Synthetic Substitution Unit: 7 Lesson: The method described below can be used to quickly evaluate a polynomial function at given values of x without the aid of technology. Example: If = x + 6x 9x 1, find f (). x-value coefficients First, write the coefficients (in order, according to descending powers) Write the x-value to be substituted up and to the left Bring down (or, copy) the leading coefficient below. Then repeat these two steps: Multiply by the x-value, and write the answer under the next coefficient to the right. Add down, and write the answer below. Continue multiplying when you move over, and adding when you move down. The final value will be the function evaluated at the given x-value. Or, in this case, f () =. Using the same function, evaluate f () and f ( 7) Use synthetic substitution to evaluate each function at the given values of x. 1) = x 6x + 1x 8 ) = x 6x + 1x 8 ) g ( x) = x + x 1x + Find f (). Find f (). Find g () ) h ( x) = x 17x 59x+ 5) h ( x) = x 17x 59x+ 6) g ( x) = x + x 1x + Find h ( ). Find h (8). 1 Find g ( ). 1, TESCCC 1/1/1 page 1 of

27 Synthetic Substitution Unit: 7 Lesson: 7) = x + 5x 1x x+ 8) = x + 5x 1x x+ Find f (). Find f ( 6). For m ( x) = x 7, find m(). To use synthetic substitution when terms of a certain power are not present, a coefficient of zero must be used to fill the place For example, the coefficients of x 7 would be,, and -7 (because the polynomial has no x term). 1 1 Continue using synthetic substitution to evaluate each polynomial. 9) = x 1x+ 16 1) a ( x) = x + x + Find f ( ). Find a (). 11) ( ) 1 7 f x = x x + x 1) y ( x) = x 17x + 16 Find f (). Find y ( ). 1, TESCCC 1/1/1 page of

28 Function Factors KEY Unit: 7 Lesson: The functions shown below are in factored form. For each, sketch the graph using a calculator window [-8, 8] by [-5, 5]. Then, tell the x-intercepts of the function, as well as its degree (quadratic, cubic, or quartic). 1) = ( x+ )( x ) ) = x( x+ )( x ) ) = x( x+ )( x ) Degree: (Quadratic) Degree: (Cubic) Degree: (Cubic) x-intercepts: x = -, x-intercepts: x = -,, x-intercepts: x = -,, ) = ( x+ )( x )( x 1) 5) = ( x+ )( x 5)( x 1) 6) = ( x+ )( x 5)( x 1)( x ) Degree: (Cubic) Degree: (Cubic) Degree: (Quartic) x-intercepts: x = -, 1, x-intercepts: x = -, 1, 5 x-intercepts: x = -, 1,, 5 7) ( = x f = x+ 9) f x) ( ) 8) ( x) ( x ) ( ) f ( x) = ( x ) ( x+ ) Degree: (Quadratic) Degree: (Cubic) Degree: (Quartic) x-intercepts: x = x-intercepts: x = -, x-intercepts: x = -, 1, TESCCC 1/1/1 page 1 of

29 Function Factors KEY Unit: 7 Lesson: 1) = ( x+ ) ( x 1)( x ) 11) = x( x+ ) ( x+ 6) 1) = 1.5x( x ) ( x+ ) Degree: (Quartic) Degree: (Quartic) Degree: (Quartic) x-intercepts: x = -, 1, x-intercepts: x = -6, -, x-intercepts: x = -,, 1) ( ) = x f x + 9 1) = ( x + 9)( x ) 15) = ( x + 1)( x )( x+ ) Degree: (Quadratic) Degree: (Cubic) Degree: (Quartic) x-intercepts: None x-intercepts: x = x-intercepts: x = -, Questions: 16) Describe the relationship between the number of factors in the function and its degree. Typically, the number of factors is the same as the degree. However, for this rule to apply, factors that are squared, or ones that contain an x, must be counted twice. 17) What is the general relationship between the factor of a function and the x-intercept related to that factor? Generally, for a number r, if (x r) is a factor of the function, the x = r is an x-intercept. 18) Graphically, what is the difference between an x-intercept caused by a single factor, (x ), and one that is generated by a factor that is squared, (x )? If (x r) is a factor of the function, then the graph crosses the x-axis at x = r. However, if (x r) is a factor of the function, then the graph touches but does not cross the x-axis at x = r. (Or, the graph is tangent to the x-axis there.) 1, TESCCC 1/1/1 page of

30 Function Factors KEY Unit: 7 Lesson: 19) Only two of the preceding functions have graphs that point down on the right (or, approach as x increases). Which ones are they? Why? # and #1, because each has a negative number as a multiplier in front of the other factors. ) True or false: The number of x-intercepts on a polynomial function is the same as the function s degree. Explain your answer. False. Generally, the number of x-intercepts is less than or equal to the degree. For example, a quadratic function (degree = ) can have two x-intercepts (see #1), one x-intercept (see #7), or none at all (see #1). 1) Without graphing, tell the x-intercepts of each function described below. A) = ( x+ 8)( x 6)( x 1) B) = x( x+ 1) ( x.7)( x.8) x = -8, 6, and 1 x =, -1,.7, and.8 ) Without the aid of a calculator, sketch the graph of each function given below. A) = ( x+ )( x 1)( x ) B) = x( x+ )( x ) C) = ( x+ )( x+ )( x 1) ) Write a factored function for each graph below. A) f(x) = x(x + )(x )(x ) B) f(x) = (x + )(x ) (x 5) 1, TESCCC 1/1/1 page of

31 Function Factors Unit: 7 Lesson: The functions shown below are in factored form. For each, sketch the graph using a calculator window [-8, 8] by [-5, 5]. Then, tell the x-intercepts of the function, as well as its degree (quadratic, cubic, or quartic). 1) = ( x+ )( x ) ) = x( x+ )( x ) ) = x( x+ )( x ) Degree: Degree: Degree: x-intercepts: x-intercepts: x-intercepts: ) = ( x+ )( x )( x 1) 5) = ( x+ )( x 5)( x 1) 6) = ( x+ )( x 5)( x 1)( x ) Degree: Degree: Degree: x-intercepts: x-intercepts: x-intercepts: 7) ( = x f = x+ 9) f x) ( ) 8) ( x) ( x ) ( ) f ( x) = ( x ) ( x+ ) Degree: Degree: Degree: x-intercepts: x-intercepts: x-intercepts: 1, TESCCC 1/1/1 page 1 of

32 Function Factors Unit: 7 Lesson: 1) = ( x+ ) ( x 1)( x ) 11) = x( x+ ) ( x+ 6) 1) = 1.5x( x ) ( x+ ) Degree: Degree: Degree: x-intercepts: x-intercepts: x-intercepts: 1) ( ) = x f x + 9 1) = ( x + 9)( x ) 15) = ( x + 1)( x )( x+ ) Degree: Degree: Degree: x-intercepts: x-intercepts: x-intercepts: Questions: 16) Describe the relationship between the number of factors in the function and its degree. 17) What is the general relationship between the factor of a function and the x-intercept related to that factor? 18) Graphically, what is the difference between an x-intercept caused by a single factor, (x ), and one that is generated by a factor that is squared, (x )? 1, TESCCC 1/1/1 page of

33 Function Factors Unit: 7 Lesson: 19) Only two of the preceding functions have graphs that point down on the right (or, approach as x increases). Which ones are they? Why? ) True or false: The number of x-intercepts on a polynomial function is the same as the function s degree. Explain your answer. 1) Without graphing, tell the x-intercepts of each function described below. A) = ( x+ 8)( x 6)( x 1) B) = x( x+ 1) ( x.7)( x.8) ) Without the aid of a calculator, sketch the graph of each function given below. A) = ( x+ )( x 1)( x ) B) = x( x+ )( x ) C) = ( x+ )( x+ )( x 1) ) Write a factored function for each graph below. A) f(x) = B) f(x) = 1, TESCCC 1/1/1 page of

34 Graphs and Factors KEY Unit: 7 Lesson: The Factor Theorem: For a polynomial function f(x), (x r) is a factor of f(x) if and only if f(r) = (or, if x = r is an x-intercept). For example, the graph shows the function = x x 11x +. Since its graph has x-intercepts at x = -,, and 5, the function can be factored as: = ( x+ )( x )( x 5) A. Use a calculator to sketch the graph of each polynomial function. Then, describe the x- intercepts, and write the function in factored form. 1) = x + x ) = x + 5x 8x 1 x-intercepts: x = -5, x-intercepts: x = -6, -1, Factored: (x + 5)(x ) Factored: (x + 6)(x + 1)(x ) ) = x 1x+ 1 ) = x x x+ 18 x-intercepts: x = -, 1, x-intercepts: x = -, (but is a double) Factored: (x + )(x 1)(x ) Factored: (x + )(x ) Squared! 5) x = + 16x 6) = x 1x + 1x+ x-intercepts: x = -,, x-intercepts: x = -1,, 5 Factored: -x (x + )(x ) Negative! Factored: (x + 1)(x )(x 5) 1, TESCCC 1/1/1 page 1 of

35 Graphs and Factors KEY Unit: 7 Lesson: B. Check to make sure your answers match the original polynomial functions by multiplying the factors you got on #1-6. Distribution. For the original example, the x-intercepts are at x = -,, and 5, so the function can be factored as: = ( x+ )( x )( x 5). Multiplying the first two factors yields: ( x + )( x ) = x x+ x 6= x + x 6 However, this quadratic trinomial still needs to be multiplied by (x 5). A few different methods are shown below. Lining Up. Multiply each Write the factors like a term by x, and traditional times each term by -5. problem. Multiply the Then combine top row by -5, then by like terms. x. Then combine like terms. 1) Factors: (x + 5)(x ) ) Factors: (x + 6)(x + 1)(x ) See student work. See student work. WORK WORK Is the answer x + x? Is the answer x + 5x 8x 1? ) Factors: (x + )(x 1)(x ) ) Factors: (x + )(x ) See student work. See student work. For example, students may forget that (x ) should be a factor twice (or, squared). WORK WORK Is the answer x 1x+ 1? Is the answer x x x + 18? 5) Factors: -x (x + )(x ) 6) Factors: (x + 1)(x )(x 5) See student work. See student work. For example, students may forget For example, students may not yet the negative sign. know to include a factor of. WORK WORK Is the answer x + 16x? Is the answer x 1x + 1x+? Did all of your factors check out when multiplied back together? Why or why not? Answers may vary. Students may have neglected to include some factors in #, 5, and 6. 1, TESCCC 1/1/1 page of

36 Graphs and Factors Unit: 7 Lesson: The Factor Theorem: For a polynomial function f(x), (x r) is a factor of f(x) if and only if f(r) = (or, if x = r is an x-intercept). For example, the graph shows the function = x x 11x +. Since its graph has x-intercepts at x = -,, and 5, the function can be factored as: = ( x+ )( x )( x 5) A. Use a calculator to sketch the graph of each polynomial function. Then, describe the x- intercepts, and write the function in factored form. 1) = x + x ) = x + 5x 8x 1 x-intercepts: x-intercepts: Factored: Factored: ) = x 1x+ 1 ) = x x x+ 18 x-intercepts: x-intercepts: Factored: Factored: 5) x = + 16x 6) = x 1x + 1x+ x-intercepts: x-intercepts: Factored: Factored: 1, TESCCC 1/1/1 page 1 of

37 Graphs and Factors Unit: 7 Lesson: B. Check to make sure your answers match the original polynomial functions by multiplying the factors you got on #1-6. Distribution. For the original example, the x-intercepts are at x = -,, and 5, so the function can be factored as: = ( x+ )( x )( x 5). Multiplying the first two factors yields: ( x + )( x ) = x x+ x 6= x + x 6 However, this quadratic trinomial still needs to be multiplied by (x 5). A few different methods are shown below. Lining Up. Multiply each Write the factors like a term by x, and traditional times each term by -5. problem. Multiply the Then combine top row by -5, then by like terms. x. Then combine like terms. 1) Factors: ) Factors: WORK WORK Is the answer x + x? Is the answer x + 5x 8x 1? ) Factors: ) Factors: WORK WORK Is the answer x 1x+ 1? Is the answer x x x 18? 5) Factors: 6) Factors: WORK WORK Is the answer x + 16x? Is the answer x 1x + 1x+? Did all of your factors check out when multiplied back together? Why or why not? 1, TESCCC 1/1/1 page of

38 Sports Puns KEY Unit: 7 Lesson: Multiply each set of factors to match them with an expanded polynomial. Then answer the riddles. 1. ( + x ) 7. (x + 7x 15)( x 1). ( x )( x+ ) 8. ( x 1)( x )(x+ 5). ( x )(x+ ) 9. ( x 1)( x+ )( x 5). ( x + 1)(x 9) 1. (x 1) ( x+ 15) 5. (x + 5)( x + x+ ) 11. ( x 6x+ 9)( x + x+ 1) 6. (x )(x + x 5) 1. ( x + x )( x x ) AC x + x x+ 15 LE x 1x + 9 AR x 5x 9 QU x 9x 58x+ 15 BA x 19x+ 15 R x + 1x+ 9 CK x 9 RS x + 1x + x+ 15 EB x + 9x 9 TE x x x + 1x+ 9 HE x + 56x 59x+ 15 TH x x 9x+ 15 Where do hair colorists sit when they go to sporting events? The Bleachers TH EB LE AC HE RS What kind of athlete gives refunds? Quarterback QU AR TE R BA CK , TESCCC 1/1/1 page 1 of 1

39 Sports Puns Unit: 7 Lesson: Multiply each set of factors to match them with an expanded polynomial. Then answer the riddles. 1. ( + x ) 7. (x + 7x 15)( x 1). ( x )( x+ ) 8. ( x 1)( x )(x+ 5). ( x )(x+ ) 9. ( x 1)( x+ )( x 5). ( x + 1)(x 9) 1. (x 1) ( x+ 15) 5. (x + 5)( x + x+ ) 11. ( x 6x+ 9)( x + x+ 1) 6. (x )(x + x 5) 1. ( x + x )( x x ) AC x + x x+ 15 LE x 1x + 9 AR x 5x 9 QU x 9x 58x+ 15 BA x 19x+ 15 R x + 1x+ 9 CK x 9 RS x + 1x + x+ 15 EB x + 9x 9 TE x x x + 1x+ 9 HE x + 56x 59x+ 15 TH x x 9x+ 15 Where do hair colorists sit when they go to sporting events? What kind of athlete gives refunds? , TESCCC 1/1/1 page 1 of 1

40 Finding a Way KEY Unit: 7 Lesson: c d b a f The graph shows the following functions: a( x) = x 9x b( x) = x c( x) = x d( x) = x = 1 x 18x 7x + 9x + x 1) Which function goes with which graph? Label the functions in the picture with a, b, c, d, or f. Here, b(x) is bold, c(x) is dashed, d(x) is dotted, and f(x) is red. ) Compare and contrast these functions. What do they have in common? How are they different? Symbolically, all the functions are cubic binomials. However, each is a scalar multiple of the other. For example, - f(x) = a(x), and a(x) = c(x). Graphically, all the functions have the same three x-intercepts (at x = -,, and ). However, each is a vertical stretch and/or reflection of the other. For example, if a(x) is reflected over the x-axis and shrunk vertically by a factor of 1/, then f(x) would result. ) Which of the given functions can be correctly factored as x ( x+ )( x )? How do you know? The function a(x) can be factored as x(x + )(x ). You can multiply the factors to show the two expressions are equal, or see that the coefficients would match up. ) How might the other functions be factored? Since each is a scalar multiple of a(x), they can be factored as k x(x + )(x ). For example, b(x) = x(x + )(x ), c(x) = x(x + )(x ), d(x) = -1x(x + )(x ), etc. 5) Match each function with one of the graphs shown. Label the functions in the picture with f, g, h, or p. = x g( x) = x h( x) =.x 1x x x p( x) = x + 1x 9 g(x) is bold, h(x) is dashed, p(x) is dotted. 6) Each of the functions has x-intercepts at x = -, -1, 1, and. This means that each one can be factored as a ( x+ )( x+ 1)( x 1)( x ) for some value of a. Find the value of a that correctly factors each function. f (x) = 1 ( x + )( x+ 1)( x 1)( x ) h (x) =. ( x + )( x+ 1)( x 1)( x ) g (x) = ( x + )( x+ 1)( x 1)( x ) p (x) = -1( x + )( x+ 1)( x 1)( x ) g f h p 1, TESCCC 1/1/1 page 1 of 5

41 Finding a Way KEY Unit: 7 Lesson: Use the information provided in each description to determine the polynomial function of least degree that has the given x-intercepts and contains the given point. Example: Find the polynomial function that passes through (, 1) with x-intercepts at x = and x = -6. First, it may help to sketch the graph: Because of the x-intercepts, two of the factors are (x + 6) and (x ). However, there may also be a numerical factor, a. Write the function as: = a( x+ 6)( x ) Because the function contains (, 1), then f ( ) = 1 (or, when x =, y = 1). Substitute these values into the equation to solve for a. f( x) = a( x+ 6)( x ) f () = a(+ 6)( ) = 1 a(6)( ) = 1 a= 1 a= 1 1 ( x) = ( x+ 6)( x So, the function is f ) To write the function as a polynomial, multiply and distribute = ( x+ 6)( x ) = ( x x+ 6x ) = ( x + x ), or = x x+ 1 7) Find the cubic polynomial function with x- intercepts at x = -,, and 6, and which also passes through (5, 1). f(x) = a(x + )(x )(x 6) Since f(5) = 1, a = -1.5 Polynomial: f(x) = -1.5x + 1.5x + 9x 18 8) Find the cubic polynomial function with x- intercepts at x = -,, and, and which also passes through (, 8). f(x) = a(x + )(x )(x ) Since f() = 8, a = ½ =.5 Polynomial: f(x) =.5x x x + 8 9) Find the quartic polynomial function with a double-root at x =, and two other x- intercepts at x = - and ; and which also passes through (1, -16). f(x) = a(x )(x )(x + ) Since f(1) = -16, a = Polynomial: f(x) = x 18x 1) Find the quartic polynomial function with a double-root at x =, and two other x- intercepts at x = - and ; and which also passes through (, -1). f(x) = a(x)(x ) (x + ) Since f() = -1, a = -/ Polynomial: f(x) = - x + x + 8 x 16 x 1, TESCCC 1/1/1 page of 5

42 Finding a Way KEY Unit: 7 Lesson: 11. Determine an appropriate representative function for each graph. Graph Function A f x x x x ( ) = B f x x x x ( ) = C f x x x x ( ) = D f x x x x ( ) = , TESCCC 1/1/1 page of 5

43 Finding a Way KEY Unit: 7 Lesson: Graph Function E f x x x x ( ) = F f x x x x ( ) = G f x x x ( ) = H f x x x ( ) = , TESCCC 1/1/1 page of 5

44 Finding a Way KEY Unit: 7 Lesson: Graph Function I =.5x +.5x + 6x J 1 f( x) = x x 5x + 1x K =.x + 1.6x.x 1x L = x + 6x 1x x 1, TESCCC 1/1/1 page 5 of 5

45 Finding a Way Unit: 7 Lesson: The graph shows the following functions: ( ) 9 a x = x x ( ) 18 b x = x ( ) 7 c x = x ( ) 9 d x = x + x 1 ( ) = f x x x x x 1. Which function goes with which graph? Label the functions in the picture with a, b, c, d, or f. ) Compare and contrast these functions. What do they have in common? How are they different? ) Which of the given functions can be correctly factored as x ( x+ )( x )? How do you know? ) How might the other functions be factored? 5) Match each function with one of the graphs shown. Label the functions in the picture with f, g, h, or p. = x g( x) = x h( x) =.x p( x) = x 1x x x + 1x ) Each of the functions has x-intercepts at x = -, -1, 1, and. This means that each one can be factored as a ( x+ )( x+ 1)( x 1)( x ) for some value of a. Find the value of a that correctly factors each function. f (x) = ( x + )( x+ 1)( x 1)( x ) h (x) = ( x + )( x+ 1)( x 1)( x ) g (x) = ( x + )( x+ 1)( x 1)( x ) p (x) = ( x + )( x+ 1)( x 1)( x ) 1, TESCCC 1/1/1 page 1 of 5

46 Finding a Way Unit: 7 Lesson: Use the information provided in each description to determine the polynomial function of least degree that has the given x-intercepts and contains the given point. Example: Find the polynomial function that passes through (, 1) with x-intercepts at x = and x = -6. First, it may help to sketch the graph: Because of the x-intercepts, two of the factors are (x + 6) and (x ). However, there may also be a numerical factor, a. Write the function as: = a( x+ 6)( x ) Because the function contains (, 1), then f ( ) = 1 (or, when x =, y = 1). Substitute these values into the equation to solve for a. = a( x+ 6)( x ) f () = a(+ 6)( ) = 1 a(6)( ) = 1 a= 1 a= 1 1 ( x) = ( x+ 6)( x So, the function is f ) To write the function as a polynomial, multiply and distribute. ( ) ( 6)( ) ( 6 ) ( 1 f x = x+ x = x x+ x = x x ), or = x + x+ 1 7) Find the cubic polynomial function with x- intercepts at x = -,, and 6, and which also passes through (5, 1). 8) Find the cubic polynomial function with x- intercepts at x = -,, and, and which also passes through (, 8). 9) Find the quartic polynomial function with a double-root at x =, and two other x- intercepts at x = - and ; and which also passes through (1, -16). 1) Find the quartic polynomial function with a double-root at x =, and two other x- intercepts at x = - and ; and which also passes through (, -1). 1, TESCCC 1/1/1 page of 5

47 Finding a Way Unit: 7 Lesson: 11. Determine an appropriate representative function for each graph. Graph Function A B C D 1, TESCCC 1/1/1 page of 5

48 Finding a Way Unit: 7 Lesson: Graph Function E F G H 1, TESCCC 1/1/1 page of 5

49 Finding a Way Unit: 7 Lesson: Graph Function I J K L 1, TESCCC 1/1/1 page 5 of 5

50 Polynomial Equations KEY Unit: 7 Lesson: The Zero Product Rule If A B C =, then either A =, B =, or C =. (x + 8)(x 1)(x 7) = x + 8 = x = -8 x 1 = x = 1 x 7 = x = 7 =. 5 When an equation is formed by setting a polynomial equal to zero, the solutions can be found by factoring the polynomial, and setting each factor equal to zero. Use this rule to help find the solutions (or roots) to each equation below. Solutions: x = {-8, 1,.5} Polynomial Equation Factored Equation Solutions (Roots) Ex. x 11x + 1= ( x 5)( x ) = x 5 =, or x = x = { 5, } 1) x + 8x + 5x 1 = ( x + 7)( x+ )( x 1) = x = {-7, -, 1} ) x + 9x 5x 1 = ( x + 11)( x+ )( x ) = x = {-5.5, -, } ) 16x 65x + = ( x 1)(x+ 1)( x )( x+ ) = x = {-, - 1, 1, } ) x + 5x 11x 55 = ( x + 5)( x 11) = x = {-5, - 11, 11} ( ) 5) x 1x 9x + 5 = ( x 6)(x 9) = x = {6, -, } (.11) 6) x 5x x + 6 = (x 9)( x + x 7) = Use the quadratic formula. x = {.5, -1+, -1 } Approx and , TESCCC 1/1/1 page 1 of

51 Polynomial Equations KEY Unit: 7 Lesson: To solve a polynomial equation when the factors are not known, graphing = x + x x the related function may help. As you know, the zeros (or x-intercepts) of the function f (x) correspond to the solutions (or roots) of the equation =. For example, solving the equation x + x x = can be accomplished by graphing = x + x x and locating the x- intercepts at x = -6, x = -1, and x =. Use this skill to help find the solutions to each equation below, as well as the related factors. Polynomial Equation Factored Equation Solutions (Roots) 7) x x 1x+ = (x + )(x )(x ) = x = {-,, } 8) 6x + 17x 1x 1= 6(x )(x + 1 )(x + ) =, or (x )(x + 1)(x + ) = x = {, - 1, -} 9) x + 5x 5x 9 = Maybe: (x+1)(x+5.656)(x 1.656) = Or: (x + 1)(x + x 9) = x {-1, , 1.656}, or, exactly: x = {-1, -+ 1, - 1 } 1) x 7x + 16x 8= Missing factor? Maybe (x 7)(???) =, Or (x 1.75)(???) = x = {1.75} 11) Half of the equations in this activity (#1, 7, 8) have only rational solutions. What does this mean? Rational solutions are those that can be written as a ratio of integers. For example, x = -5.5 can be written as x = -11/. This is a rational solution because -11 and are integers. 1) Other equations in this activity (# 6, 9) have irrational solutions. What does this mean? Irrational solutions cannot be written as ratios of integers. Examples include those that involve square root values such as x = 11 or x = Think About It: Besides rational and irrational values, can yet another type of solution exist? Students might remember that solutions can be imaginary or complex. 1, TESCCC 1/1/1 page of

52 Polynomial Equations Unit: 7 Lesson: The Zero Product Rule If A B C =, then either A =, B =, or C =. (x + 8)(x 1)(x 7) = x + 8 = x = -8 x 1 = x = 1 x 7 = x = 7 =. 5 When an equation is formed by setting a polynomial equal to zero, the solutions can be found by factoring the polynomial, and setting each factor equal to zero. Use this rule to help find the solutions (or roots) to each equation below. Solutions: x = {-8, 1,.5} Polynomial Equation Factored Equation Solutions (Roots) Ex. x 11x + 1= ( x 5)( x ) = x 5 =, or x = x = { 5, } 1) ( x + 7)( x+ )( x 1) = ) ( x + 11)( x+ )( x ) = ) ( x 1)(x+ 1)( x )( x+ ) = ) ( x + 5)( x 11) = 5) ( x 6)(x 9) = 6) (x 9)( x + x 7) = 1, TESCCC 1/1/1 page 1 of

53 Polynomial Equations Unit: 7 Lesson: To solve a polynomial equation when the factors are not known, graphing = x + x x the related function may help. As you know, the zeros (or x-intercepts) of the function f (x) correspond to the solutions (or roots) of the equation =. For example, solving the equation x + x x = can be accomplished by graphing = x + x x and locating the x- intercepts at x = -6, x = -1, and x =. Use this skill to help find the solutions to each equation below, as well as the related factors. Polynomial Equation Factored Equation Solutions (Roots) 7) x x 1x+ = 8) 6x + 17x 1x 1= 9) x + 5x 5x 9 = 1) x 7x + 16x 8= 11) Half of the equations in this activity (#1, 7, 8) have only rational solutions. What does this mean? 1) Other equations in this activity (# 6, 9) have irrational solutions. What does this mean? Think About It: Besides rational and irrational values, can yet another type of solution exist? 1, TESCCC 1/1/1 page of

54 Solution Scramble KEY Unit: 7 Lesson: b± b ac Use the quadratic formula ( x = ) to help find all the solutions to each factored a equation. The solutions for each equation will correspond to letters below that can be unscrambled to spell four common words. Equation Solutions Word 1. (x 7)(6x 7x ) = x = 7 x = 5 x = TT BU ON BUTTON. (7x )(x 7x ) = x = 7 x = 5 x = IO CUR US CURIOUS. (x + 5)( x + x 1) = 5 x = x = x = EST LER WR WRESTLER. ( x 8x+ 5)( x x+ ) = IL NAI LIO RE MILLIONAIRE Solution Bank: 5 x = 7 x = x = x = x = IO ECK CUR FUR US x = x = x = x = x = 15 LER MIL MAY ON EST x = + 11 x = x = x = x = LIO NED CHA WR NAI x = x = 5 x = 11 x = x = SHU TT RE ASE BU BONUS: What do all the unscrambled words have in common? All the words appeared in titles of films nominated for Academy Awards in 8 (Slumdog Millionaire, The Curious Case of Benjamin Button, The Wrestler). 1, TESCCC 1/1/1 page 1 of 1

55 Solution Scramble Unit: 7 Lesson: b± b ac Use the quadratic formula ( x = ) to help find all the solutions to each factored a equation. The solutions for each equation will correspond to letters below that can be unscrambled to spell four common words. Equation Solutions Word 1. (x 7)(6x 7x ) =. (7x )(x 7x ) =. (x + 5)( x + x 1) =. ( x 8x+ 5)( x x+ ) = Solution Bank: 5 x = 7 x = x = x = x = IO ECK CUR FUR US x = x = x = x = x = 15 LER MIL MAY ON EST x = + 11 x = x = x = x = LIO NED CHA WR NAI x = x = 5 x = 11 x = x = SHU TT RE ASE BU BONUS: What do all the unscrambled words have in common? 1, TESCCC 1/1/1 page 1 of 1

56 The Missing Factor KEY Unit: 7 Lesson: 9775 =? What operation is used to find the missing factor? Division Explain the steps in doing the arithmetic necessary to compute the solution to this equation without the aid of a calculator. Ask, What can I multiply by to get close to 97? (It s. Write this above.) Multiply by. Write this product under the 97 and subtract to get 5. Bring down the 7. Then repeat: Ask, What can I multiply by to get close to 57? (It s. Write this above.) Multiply by. Write this product under the 57 and subtract to get 11. Bring down the 5. Repeat again, etc. x 5x 5x 9 ( x 1) (? ) B A x + x 9 x + 1 x 5x 5x 9 (x + x ) C D E x 5x (x + x) 9x 9 ( 9x 9) To find missing factors in polynomial problems, we can use a similar process called polynomial long division. Follow these steps: A. Think, What can I multiply (x + 1) by so that the answer looks like x +? Write this term above division bar. B. Multiply these expressions and write them underneath. C. Subtract the two expressions (or, add the opposite). D. Write the answer below. E. Bring down the next term. Then, repeat with the resulting expression. 1) x + x 5 x x 15 (x 5x) 6x 15 (6x 15) By multiplying the factors, we know that ( x 5)( x ) equals x x 15. Show, using polynomial long division, that: 1) (x x 15) (x 5) x, and ) x 5 x + x x 15 (x + 6x) 5x 15 (-5x 15) ) (x x 15) ( x ) x 5 1, TESCCC /6/1 page 1 of

57 The Missing Factor KEY Unit: 7 Lesson: Use polynomial long division to find the missing factors. ) x 17x 1x ( x 5) (? ) ) x x 1x 19x 1 ( x ) (? ) x 5 x 17x 1x x x x 1x 19x 1 Answer: x + 7x Answer: x + x 5x + 5) x 9x x 1x 15 (x ) (? ) 6) x 7x 16x 8 (x 7) (? ) x x 9x x 1x 15 x 7 x 7x 16x 8 Answer: x 6x + 8x 5 Answer: x + 1, TESCCC /6/1 page of

58 The Missing Factor KEY Unit: 7 Lesson: 7) Consider the function x 7x 5x. A) For which one of the given values of x is? Circle one. Then use synthetic substitution to check your answer. x = x = x = 5 B) Based on your answer to part A, which of the following would be a factor of f (x)? Circle one. (x ) (x ) (x 5) C) Using the factor from part B, find the other factor using polynomial long division. Answer: x + 11x 6 8) Consider the function g ( x) x x x 8. A) For which one of the given values of x is g ( x)? Circle one. Then use synthetic substitution to check your answer. x = - x = x = -8 B) Based on your answer to part A, which of the following would be a factor of g (x)? Circle one. (x ) (x + ) (x 8) C) Using the factor from part B, find the other factor using polynomial long division. Answer: x x x + 11x 6 x x 7x 5x (x x ) - 11x 5x (11x x) x + (-6x + ) x x + x + x x x 8 (x + x ) -x x (-x 6x) x + 8 (x + 8) 9) LOOK: What similarities exist between synthetic substitution and the polynomial long division? The numbers below the line in the substitution are the same as the coefficients of the other factor. 1, TESCCC /6/1 page of

59 The Missing Factor KEY Unit: 7 Lesson: When synthetic substitution is used to check a solution to a polynomial equation of the form p(x) =, a great deal of extra information is revealed. Specifically, the numbers under the substitution table are the coefficients (in order) to the missing factor! Since this missing factor is one degree less that the original polynomial, it is sometimes called the depressed polynomial. p( x) x x x 18 Verify that x = - is a solution to p(x) = Coefficients of the other factor The zero at the end means that p(-) =. So (x + ) is a factor of p(x). And the other factor must be x 5x 9 Or, p ( x) x x x 18 factors as ( x )( x 5x 9). For each equation below, one solution is provided. Use synthetic substitution both to check this solution and to write the factors of the polynomial in the equation. From the depressed polynomial find the other solutions, using the quadratic formula. 1) x 5x 19x Factored equation: One solution: x = (x )(x x 1) = Solve x x 1 = by factoring, or using the quadratic formula with a =, b = -1, and c = -1. Other solutions: x =.5 x = - 11) x x 9x 18 Factored equation: One solution: x = (x )( x 11x + 6) = Solve x 11x + 6 = by factoring, or using the quadratic formula with a =, b = -11, and c = 6. Other solutions: x = (again) x = 1) x x 17x Factored equation: One solution: x = - (x + )( x x 5) = Solve x x 5 = using the quadratic formula with a = 1, b = -, and c = 5. Other solutions: x = 9 x.196 x = 9 x , TESCCC /6/1 page of

60 The Missing Factor Unit: 7 Lesson: 9775 =? What operation is used to find the missing factor? 9775 Explain the steps in doing the arithmetic necessary to compute the solution to this equation without the aid of a calculator. x 5x 5x 9 ( x 1) (? ) To find missing factors in polynomial problems, we can use a similar process called polynomial long division. x + 1 x 5x 5x 9 Follow these steps: A. Think, What can I multiply (x + 1) by so that the answer looks like x +? Write this term above division bar. B. Multiply these expressions and write them underneath. C. Subtract the two expressions (or, add the opposite). D. Write the answer below. E. Bring down the next term. Then, repeat with the resulting expression. 1) x 5 x x 15 By multiplying the factors, we know that ( x 5)( x ) equals x x 15. Show, using polynomial long division, that: ) x + x x 15 1) (x x 15) (x 5) x, and ) (x x 15) ( x ) x 5 1, TESCCC 1/1/1 page 1 of

61 The Missing Factor Unit: 7 Lesson: Use polynomial long division to find the missing factors. ) x 17x 1x ( x 5) (? ) ) x x 1x 19x 1 ( x ) (? ) x 5 x 17x 1x x x x 1x 19x 1 5) x 9x x 1x 15 (x ) (? ) 6) x 7x 16x 8 (x 7) (? ) x x 9x x 1x 15 x 7 x 7x 16x 8 1, TESCCC 1/1/1 page of

62 The Missing Factor Unit: 7 Lesson: 7) Consider the function x 7x 5x. A) For which one of the given values of x is? Circle one. Then use synthetic substitution to check your answer. x = x = x = B) Based on your answer to part A, which of the following would be a factor of f (x)? Circle one. (x ) (x ) (x 5) C) Using the factor from part B, find the other factor using polynomial long division. 8) Consider the function g ( x) x x x 8. A) For which one of the given values of x is g ( x)? Circle one. Then use synthetic substitution to check your answer. x = - x = x = B) Based on your answer to part A, which of the following would be a factor of g (x)? Circle one. (x ) (x + ) (x 8) C) Using the factor from part B, find the other factor using polynomial long division. 9) LOOK: What similarities exist between synthetic substitution and the polynomial long division? 1, TESCCC 1/1/1 page of

63 The Missing Factor Unit: 7 Lesson: When synthetic substitution is used to check a solution to a polynomial equation of the form p(x) =, a great deal of extra information is revealed. Specifically, the numbers under the substitution table are the coefficients (in order) to the missing factor! Since this missing factor is one degree less that the original polynomial, it is sometimes called the depressed polynomial. p( x) x x x 18 Verify that x = - is a solution to p(x) = Coefficients of the other factor The zero at the end means that p(-) =. So (x + ) is a factor of p(x). And the other factor must be x 5x 9 Or, p ( x) x x x 18 factors as ( x )( x 5x 9). For each equation below, one solution is provided. Use synthetic substitution both to check this solution and to write the factors of the polynomial in the equation. From the depressed polynomial find the other solutions, using the quadratic formula. 1) x 5x 19x Factored equation: One solution: x = (x )( ) = Other solutions: 11) x x 9x 18 Factored equation: One solution: x = (x )( ) = Other solutions: 1) x x 17x Factored equation: One solution: x = - (x + )( ) = Other solutions: 1, TESCCC 1/1/1 page of

64 The Complex Nature of Things KEY Unit: 7 Lesson: Given a cubic polynomial function and one of its zeros (or, x-intercepts), find the other zeros using synthetic factoring and the quadratic formula. 1) = x + x 1x + 8 Factored function & equation: One zero: f ( 1) = f(x) = (x 1)(x + 5x 8) = Solve x x 8 = using the quadratic formula with a =, b = 5, and c = Other zeros: x = 1 (again) x = ) = x + x 6x Factored function & equation: One zero: f ( 5) = f(x) = (x + 5)(x x 6) = Solve x x 6 = using the quadratic formula with a = 1, b = -, and c = Other zeros: x = + 1 x = 1 ) = x 6x + 1x 8 Factored function & equation: One zero: f ( ) = f(x) = (x )(x x + ) = Solve x x + = using the quadratic formula with a = 1, b = -, and c =. Because of a negative -8 8 radicand, solutions involve imaginary numbers. 1 - Other zeros: x = 1 + i x = 1 i ) Interpret your results by graphing the polynomial functions from above and analyzing their x- intercepts. = x + x 1x + 8 = x + x 6x = x 6x + 1x 8 The function has a doubleroot at x = 1 The function has three distinct x-intercepts The function has only one x- intercept. 1, TESCCC //1 page 1 of

65 The Complex Nature of Things KEY Unit: 7 Lesson: The Fundamental Theorem of Algebra If a polynomial function p (x) is of degree n, then the related equation p ( x) = must have n (or, the same number of) solutions. In other words Quadratic equations (of degree ) must have solutions (roots), Cubic equations (of degree ) must have solutions (roots), Quartic equations (of degree ) must have solutions (roots), and so on. Algebraic Considerations Graphical Interpretations Some solutions can count more than once, Real solutions to p ( x) = show up as locations (such as a double root). where the graph of p (x) intersects the x-axis; In these cases, we say that the solution (or root) has multiplicity greater than 1. At a double root, the graph will touch (but not cross) the x-axis. Some solutions involve non-real numbers Non-real zeros of p (x) do NOT (with imaginary values). And, when non-real roots are present, they must come in conjugate pairs. show up as x-intercepts. Instead, they create bends in the graph that prevent it from crossing the x-axis. Sketch the graph of each function based on the description of its zeros. 5) A quadratic function with one real zero of multiplicity (or, a double root). 8) A cubic function with one real zero and two non-real zeros. 6) A quadratic function with a negative leading coefficient and two non-real zeros. 9) A quartic ( th - degree) function with two distinct real zeros and one double root. 7) A cubic function with three distinct positive real zeros. 1) A quartic function with two distinct real zeros and two nonreal zeros. 1, TESCCC //1 page of

66 The Complex Nature of Things KEY Unit: 7 Lesson: For each function graphed, describe the number and nature of its zeros (how many, and what kind). 11) A cubic function has zeros. This has one negative zero, and a positive zero of multiplicity (double root). 1) A quartic function has zeros. This has one real zero of multiplicity (double root), and two nonreal zeros. This one has a negative leading coefficient. 1) A quartic function has zeros. This has non-real zeros (two pairs). This one has a negative leading coefficient. 1) A 5 th degree function has 5 zeros. This has real zeros and non-real zeros. Each cubic function below has only one real zero. Identify this zero from the graph (or using a calculator). Next, use synthetic substitution both to check this zero and to write the factors of the polynomial. From the depressed factor, find the remaining complex zeros of the function. 15) = x x 1x + Real zero: x = x = ± 6 ( 6) (1)(1) 6± 6± i x = = Factored function: f(x) = (x + )(x 6x + 1) Complex zeros: x = + i, i 16) = x 11x + 7x 87 Real zero: x = 1.5 = x = ± 8 ( 8) ()(58) 8 ± 8 ± i x = = Factored function: f(x) = (x 1.5)(x 8x + 58) Complex zeros: x = + 5i, 5i Function could also be factored as f(x) = (x )(x x + 9) 1, TESCCC //1 page of

67 The Complex Nature of Things KEY Unit: 7 Lesson: Write a polynomial function that has the given irrational or complex zeros. When multiplying, remember that i = -1. Sample: Write a polynomial function with zeros of, i, and -i. (Think: If z is a zero, then (x z) is a factor.) ( x )( x i)( x + i) = ( x )( x + ix ix 16 i ) ( )( 16( 1)) ( )( 16) = x x = x x + = x x + 16x 8 So, the function = x x + 16x 8 has zeros of, i, and -i. 17) Write a cubic polynomial function with zeros of 5, 7, and 7. f(x) = (x 5)(x 7 )(x + 7 ) = (x 5)(x 7) = x 5x 7x ) Write a cubic polynomial function with zeros of 5, +, and. f(x) = (x 5)(x )(x + ) = (x 5)(x 8x + 1) = x 1x + 5x 65 19) Write a cubic polynomial function with zeros of 1, i, and -i. f(x) = (x 1)(x i)(x + i) = (x 1)(x + 9) = x x + 9x 9 ) Write a cubic polynomial function with zeros of -, 5 + i, and 5 i. f(x) = (x + )(x 5 i)(x 5 + i) = (x + )(x 1x + 9) = x 6x 11x , TESCCC //1 page of

68 The Complex Nature of Things Unit: 7 Lesson: Given a cubic polynomial function and one of its zeros (or, x-intercepts), find the other zeros using synthetic factoring and the quadratic formula. 1) = x + x 1x + 8 Factored function & equation: One zero: f ( 1) = f(x) = (x 1)( ) = Other zeros: ) = x + x 6x Factored function & equation: One zero: f ( 5) = f(x) = (x + 5)( ) = Other zeros: ) = x 6x + 1x 8 Factored function & equation: One zero: f ( ) = f(x) = (x )( ) = Other zeros: ) Interpret your results by graphing the polynomial functions from above and analyzing their x- intercepts. = x + x 1x + 8 = x + x 6x = x 6x + 1x 8 1, TESCCC //1 page 1 of

69 The Complex Nature of Things Unit: 7 Lesson: The Fundamental Theorem of Algebra If a polynomial function p (x) is of degree n, then the related equation p ( x) = must have In other words Quadratic equations (of degree ) must have, Cubic equations (of degree ) must have, Quartic equations (of degree ) must have, and so on. Algebraic Considerations Graphical Interpretations Some solutions can count more than once, Real solutions to p ( x) = show up as locations (such as a ). where the graph of p (x) intersects the x-axis; In these cases, we say that the solution (or root) has greater than 1. At a, the graph will (but not cross) the x-axis. Some solutions involve numbers zeros of p (x) do NOT (with values). And, when roots are present, they must come in. show up as. Instead, they create in the graph that prevent it from. Sketch the graph of each function based on the description of its zeros. 5) A quadratic function with one real zero of multiplicity (or, a double root). 8) A cubic function with one real zero and two non-real zeros. 6) A quadratic function with a negative leading coefficient and two non-real zeros. 9) A quartic ( th - degree) function with two distinct real zeros and one double root. 7) A cubic function with three distinct positive real zeros. 1) A quartic function with two distinct real zeros and two nonreal zeros. 1, TESCCC //1 page of

70 The Complex Nature of Things Unit: 7 Lesson: For each function graphed, describe the number and nature of its zeros (how many, and what kind). 11) 1) 1) 1) Each cubic function below has only one real zero. Identify this zero from the graph (or using a calculator). Next, use synthetic substitution both to check this zero and to write the factors of the polynomial. From the depressed factor, find the complex zeros of the function. 15) = x x 1x + Real zero: Factored function: Complex zeros: 16) = x 11x + 7x 87 Real zero: Factored function: Complex zeros: 1, TESCCC //1 page of

71 The Complex Nature of Things Unit: 7 Lesson: Write a polynomial function that has the given irrational or complex zeros. When multiplying, remember that i = -1. Sample: Write a polynomial function with zeros of, i, and -i. (Think: If z is a zero, then (x z) is a factor.) ( x )( x i)( x + i) = ( x )( x + ix ix 16 i ) ( )( 16( 1)) ( )( 16) = x x = x x + = x x + 16x 8 So, the function = x x + 16x 8 has zeros of, i, and -i. 17) Write a cubic polynomial function with zeros of 5, 7, and 7. 18) Write a cubic polynomial function with zeros of 5, +, and. 19) Write a cubic polynomial function with zeros of 1, i, and -i. ) Write a cubic polynomial function with zeros of -, 5 + i, and 5 i. 1, TESCCC //1 page of

72 Exact Zeros KEY Unit: 7 Lesson: Objective: Find the exact values of all the zeros of a polynomial function. Step One: Use technology to find the rational zeros of the polynomial function. Example 1: Find the rational zeros of x 7x 9x 7x 9. A) From the Graph With the graph on a calculator, locate the zeros. Left Bound Right Bound Zero (x-intercept) Left Bound and Right Bound mean to move to either side of the zero. Some zeros are simple to spot using the calculator s graph, such as integers (like x = ). Others are not so obvious. For example, the value shown is probably not a rational zero. B) From the Table When zeros are hard to identify from the graph, a table can provide more exact or detailed information. Zero between Zero between Zero between Because the y-values switch from negative to positive, the table indicates that there are zeros between x = -5 and -, between x = -1 and, and between x = and 1. Zero between Zero between To see if any of these are rational zeros, change the increment to a smaller value. Tbl =.5 Here, however, we still find no zeros. Zero Zero between Sometimes an even smaller increment is needed to see the rational zeros. Tbl =.5 Here, a rational zero occurs at x.75. C) From the List of Possibilities A special rule helps you narrow down the possibilities for rational zeros of a polynomial function. The Rational Root Theorem says that the only possible rational zeros must be of the form where p indicates possible factors ( ) of the function s constant term and q indicates possible factors ( ) of the function s leading coefficient. p, q 1, TESCCC /6/1 page 1 of

73 Exact Zeros KEY Unit: 7 Lesson: Use the Rational Root Theorem to list the possible rational zeros of each partial polynomial. x 9 p = Factors of 9: 1,, 9 q = Factors of : 1,, Zeros? 1,, 9,,,,,, Table Tip: To catch all the rational zeros in the table (and not skip over any of them), set the increment to Tbl = 1/a, where a is the leading coefficient of the polynomial function. This way, all of the possible rational zeros will be listed (as long as a is an integer). Step Two: Use synthetic substitution & factoring to find the other zeros of the polynomial function (such as irrational and non-real zeros). Example : Find the remaining zeros of x 7x 9x 7x 9, given that f ( ) and f ( ). Method: Run the known rational zeros through synthetic substitution until a quadratic polynomial results. Then use the quadratic formula to find the zeros of this depressed polynomial Use the coefficients of x 7x 9x 7x The final answer indicates that x = is a zero of f (x). With the depressed polynomial, the function factors into: f x x x x x ( ) ( )( 19 8 ) Use the coefficients of the depressed (cubic) polynomial with the other rational root (x = ) A second depressed (quadratic) polynomial results, indicating that the original function can factor into: ( x )( x )(x 16x ) x 16x 16 (16) ()( ) x () The final solutions come from setting this depressed (quadratic) polynomial equal to zero. Here, use a =, b = 16, and c = -. In this case, the final answer can simplify to x 5. So, exactly, the four zeros of the polynomial are: x = {,, 5, 5 } 1, TESCCC /6/1 page of

74 Exact Zeros KEY Unit: 7 Lesson: Find the exact values for all of the zeros of each function. 1) x x 11x A) Use the Rational Root Theorem to list the possible rational zeros of the function. B) Use a calculator to determine which of these values are, in fact, zeros of the function. C) Describe the number (how many) and nature (what kind) of the zeros this polynomial function. ) x x x 8x 6 A) Use the Rational Root Theorem to list the possible rational zeros of the function. B) Use a calculator to determine which of these values are, in fact, zeros of the function. C) Describe the number (how many) and nature (what kind) of the zeros this polynomial function. 1,,, 5, 6, 1, 15, x = {-,, 5} Cubic function has zeros. Here, there are three distinct rational zeros. 1,,, 8, 16,, 6 x = {-1, -,, } Quartic function has zeros. Here, there are rational zeros at x = -1 and -, and a zero of multiplicity (double root) at x =. ) x 1x 5x A) Use the Rational Root Theorem to list the possible rational zeros of the function. B) Use a calculator to determine which of these values are, in fact, zeros of the function ,,, 5, 6, 1, 15,,,,, x = 5 =.5 C) In the space below, use synthetic substitution (and the depressed polynomial) to find the other irrational or non-real zeros of this function. This cubic function has three district, real zeros x 16x + 1 = x x 16 1 x x = 1 One of the zeros is rational: x = 5 The other two are irrational: x = + 1, and x = 1 1, TESCCC /6/1 page of

75 Exact Zeros KEY Unit: 7 Lesson: Find the exact values for all of the zeros of each function. ) x 5x 19x 6x A) Use a calculator to find the two rational zeros of this function. x = - and x = B) Use synthetic substitution (twice) to find the (depressed) quadratic factor of this function ) x x x x 16 A) Use a calculator to find the two rational zeros of this function. x = and x = B) Use synthetic substitution (twice) to find the (depressed) quadratic factor of this function / Depressed Quadratic Factor: x 7x + Depressed Quadratic Factor: x 6x + 6 C) From this quadratic factor, find the other irrational or non-real zeros x 7 7 x Zeros: X = {-,, 7 7, 7 7 } C) From this quadratic factor, find the other irrational or non-real zeros. x 6x + 6 = 6 6 x 6 6 6i x 6 x 1 i Zeros: X = {,, 1+i, 1 i} 1, TESCCC /6/1 page of

76 Exact Zeros Unit: 7 Lesson: Objective: Find the exact values of all the zeros of a polynomial function. Step One: Use technology to find the rational zeros of the polynomial function. Example 1: Find the rational zeros of x 7x 9x 7x 9. A) From the Graph With the graph on a calculator, locate the zeros. Left Bound Right Bound Zero (x-intercept) Left Bound and Right Bound mean to move to either side of the zero. Some zeros are simple to spot using the calculator s graph, such as integers (like x = ). Others are not so obvious. For example, the value shown is probably not a zero. B) From the Table When zeros are hard to identify from the graph, a table can provide more exact or detailed information. Zero between Zero between Zero between Because the y-values switch from negative to positive, the table indicates that there are zeros between x = -5 and -, between x = -1 and, and between x = and 1. Zero between Zero between To see if any of these are rational zeros, change the increment to a smaller value. Tbl = Here, however, we still find no zeros. Zero Zero between Sometimes an even smaller increment is needed to see the rational zeros. Tbl = Here, a rational zero occurs at x.75. C) From the List of Possibilities A special rule helps you narrow down the possibilities for rational zeros of a polynomial function. The Rational Root Theorem says that the only possible rational zeros must be of the form where p indicates and q indicates p, q 1, TESCCC /6/1 page 1 of

77 Exact Zeros Unit: 7 Lesson: Use the Rational Root Theorem to list the possible rational zeros of each partial polynomial. ) x 9 p = Factors of : q = Factors of : Zeros? Table Tip: To catch all the rational zeros in the table (and not skip over any of them), set the increment to Tbl = 1/a, where a is the leading coefficient of the polynomial function. This way, all of the possible rational zeros will be listed (as long as a is an integer).. Step Two: Use synthetic substitution & factoring to find the other zeros of the polynomial function (such as irrational and complex zeros). Example : Find the other zeros of x 7x 9x 7x 9, given that f ( ) and f ( ). Method: Run the known rational zeros through synthetic substitution until a quadratic polynomial results. Then use the quadratic formula to find the zeros of this depressed polynomial Use the coefficients of x 7x 9x 7x 9. The final answer indicates that x = is a zero of f (x). With the depressed polynomial, the function factors into: f x x x x x ( ) ( )( 19 8 ) Use the coefficients of the depressed (cubic) polynomial with the other rational root (x = ) A second depressed (quadratic) polynomial results, indicating that the original function can factor into: ( x )( x )(x 16x ) x 16x 16 (16) ()( ) x () The final solutions come from setting this depressed (quadratic) polynomial equal to zero. Here, use a =, b = 16, and c = -. In this case, the final answer can simplify to x 5. So, exactly, the four zeros of the polynomial are: x = {,, 5, 5 } 1, TESCCC /6/1 page of

78 Exact Zeros Unit: 7 Lesson: Find the exact values for all of the zeros of each function. 1) x x 11x A) Use the Rational Root Theorem to list the possible rational zeros of the function. B) Use a calculator to determine which of these values are, in fact, zeros of the function. C) Describe the number (how many) and nature (what kind) of the zeros this polynomial function. ) x x x 8x 6 A) Use the Rational Root Theorem to list the possible rational zeros of the function. B) Use a calculator to determine which of these values are, in fact, zeros of the function. C) Describe the number (how many) and nature (what kind) of the zeros this polynomial function. ) x 1x 5x A) Use the Rational Root Theorem to list the possible rational zeros of the function. B) Use a calculator to determine which of these values are, in fact, zeros of the function. C) In the space below, use synthetic substitution (and the depressed polynomial) to find the other irrational or non-real zeros of this function. 1, TESCCC /6/1 page of

79 Exact Zeros Unit: 7 Lesson: Find the exact values for all of the zeros of each function. ) x 5x 19x 6x A) Use a calculator to find the two rational zeros of this function. 5) x x x x 16 A) Use a calculator to find the two rational zeros of this function. B) Use synthetic substitution (twice) to find the (depressed) quadratic factor of this function B) Use synthetic substitution (twice) to find the (depressed) quadratic factor of this function C) From this quadratic factor, find the other irrational or non-real zeros. C) From this quadratic factor, find the other irrational or non-real zeros. 1, TESCCC /6/1 page of

80 Equation Solver KEY Unit: 7 Lesson: Solve each equation for its exact roots by completing the following steps: A) By adding or subtracting terms on either side, rewrite the equation in the form p(x) =, where p(x) is a polynomial. B) Find the rational zeros of p(x) by finding possible roots using the rational root theorem and using technology to verify which are actual roots. C) Find any other (irrational or non-real) zeros using synthetic substitution and factoring or the quadratic formula on the depressed polynomial. 1) x + 7x + x = ) x + 1= x + 17x x + 7x + x = x = { 1, -, -} (x = - is a double root) x x 17x+ 1= x = {, , } ) x + 8x + = 11x + 1x ) 5x + 18x + 1x = x + 8 x + 8x 11x 1x + = x = { -,, + 19, 19 } 5x + x + 18x + 1x 8= x = { -1, 5, i, -i } 1, TESCCC 1/1/1 page 1 of 1

81 Equation Solver Unit: 7 Lesson: Solve each equation for its exact roots by completing the following steps: A) By adding or subtracting terms on either side, rewrite the equation in the form p(x) =, where p(x) is a polynomial. B) Find the rational zeros of p(x) by finding possible roots using the rational root theorem and using technology to verify which are actual roots. C) Find any other (irrational or non-real) zeros using synthetic substitution and factoring or the quadratic formula on the depressed polynomial. 1) x + 7x + x = ) x + 1= x + 17x ) x + 8x + = 11x + 1x ) 5x + 18x + 1x = x + 8 1, TESCCC 1/1/1 page 1 of 1

82 Do-It-Yourself Equations Unit: 7 Lesson: Objective: Create (or make up) a fourth-degree polynomial equation with rational coefficients that a classmate will solve. Guidelines: Only Two Rational Even though a fourth-degree equation has four solutions, your equation can only have two rational roots. This means that the other two solutions must either be irrational (with square roots) or non-real (with i terms). Remember irrational and non-real factors must occur in conjugate pairs in order for the coefficients to all be rational. Steps: 1) Pick four values to be solutions. Write them in the boxes below. Rational Solutions Irrational or Complex Solutions Pick easy integers, like, 5, -8, These must come in a conjugate pair, such as etc. + 7 and 7, or 1+ i and 1 i. ) Write the factors of the polynomial function having these zeros. (x ) (x ) (x ) (x ) ) Multiply the factors to create a polynomial. ) Write the final polynomial in the blank below (and sign your name, too). Then, separate the equation from the rest of the page by cutting along the dotted line. This is the equation you will give to a classmate to solve. Equation: = From 1, TESCCC 1/1/1 page 1 of 1

83 Polynomial Algebra KEY Unit: 7 Lesson: From the answer bank, find a function in factored form to match each graph. 1) E ) F Possible Answers: ) C ) H A. ( x )( x ) B. ( x ) ( x ) C. f( x) x( x )( x ) D. f x x x ( ) ( ) ( ) E. x( x )( x ) F. x( x ) ( x ) G. H. f x x x x ( ) ( ) ( ) f( x) x( x )( x ) Sketch a graph to match each of the functions described below. 5) 6) 7) 8) A quadratic function with two non-real zeros A cubic function with one real zero and two non-real zeros A quartic ( th -degree) function with real zeros and two non-real zeros A quartic function with one real double root and two non-real zeros Describe the number and nature of all the roots of each 5 th -degree polynomial function below. 9) 1) Three distinct real zeros, and one zero of multiplicity (double root). Three distinct real zeros, and two non-real zeros. The leading coefficient is negative. 1, TESCCC /6/1 page 1 of

84 Polynomial Algebra KEY Unit: 7 Lesson: 11) Write a cubic polynomial function to match the graph shown below. f(x) = a (x + )(x ) Since f() = -16, a = -½ f(x) = -.5(x + )(x ) f(x) = -.5 (x 1x + 16) f(x) = -.5x + 6x 8 1) Write a quadratic polynomial function with zeros of + 5i and 5i. f(x) = (x 5i)(x + 5i) f(x) = x 6x + 6x 19x x 1 (x ) (???) x x 17x 1 ( x ) (???) 1) Find the missing factor in the item above using polynomial long division. x 6x 19x Answer: x 5x + x 1 1) Find the missing factor in the item above by identifying the depressed polynomial from synthetic substitution Answer: x + 5x ) f(x) = x + 11x + x - 15 A) Use the Rational Root Theorem to list the possible rational zeros of this polynomial function. B) Use a calculator to determine which of these values are, in fact, rational zeros of f(x) ,, 5, 15,,,, x = { } 16) x x x 7x 18 A) Use the Rational Root Theorem to list the possible rational zeros of this polynomial function. B) Use a calculator to determine which of these values are, in fact, rational zeros of f(x) ,,, 6, 9, 18,,, x = {-, -, ½, } 1, TESCCC /6/1 page of

85 Polynomial Algebra KEY Unit: 7 Lesson: Determine the exact values of all the zeros of each polynomial function. Irrational zeros should be left in simplified radical form, and complex zeros should be written in the form a + bi. 17) x 1x 5x 1 18) x x 9x 79x 1 x = {, 7, 7 } x = { -5,, + i, i } 19) x 7x 8x 8 ) x 11x 5x 5x 5 x = { 7, i, -i } x = { -1,.5, 1 + 6, 1 6 } 1, TESCCC /6/1 page of

86 Polynomial Algebra PI Unit: 7 Lesson: From the answer bank, find a function in factored form to match each graph. 1) ) Possible Answers: ) ) A. ( x )( x ) B. ( x ) ( x ) C. f( x) x( x )( x ) D. ( x ) ( x ) E. x( x )( x ) F. x( x ) ( x ) G. H. x( x ) ( x ) x( x )( x ) Sketch a graph to match each of the functions described below. 5) 6) 7) 8) A quadratic function with two non-real zeros A cubic function with one real zero and two non-real zeros A quartic ( th -degree) function with real zeros and two non-real zeros A quartic function with one real double root and two non-real zeros Describe the number and nature of all the roots of each 5 th -degree polynomial function below. 9) 1) 1, TESCCC 1/1/1 page 1 of

87 Polynomial Algebra PI Unit: 7 Lesson: 11) Write a cubic polynomial function to match the graph shown below. 1) Write a quadratic polynomial function with zeros of + 5i and 5i. 6x 19x x 1 (x ) (???) x x 17x 1 ( x ) (???) 1) Find the missing factor in the item above using polynomial long division. 1) Find the missing factor in the item above by identifying the depressed polynomial from synthetic substitution. x 6x 19x x ) f(x) = x + 11x + x - 15 A) Use the Rational Root Theorem to list the possible rational zeros of this polynomial function. B) Use a calculator to determine which of these values are, in fact, rational zeros of f(x). 16) x x x 7x 18 A) Use the Rational Root Theorem to list the possible rational zeros of this polynomial function. B) Use a calculator to determine which of these values are, in fact, rational zeros of f(x). 1, TESCCC 1/1/1 page of

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