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Overview

Chapter I expand the students’ ability to use polynomials to represent and solve problems reflecting real-world situations while focusing on symbolic and graphical patterns.

The study of the properties and graphs of polynomial functions is useful to scientists, astronomers, physicists and chemists in the field of scientific research. These properties are useful in making satellite dishes, car headlights, radio telescopes and reflecting telescopes.

The ideas and skills learned in this chapter will help the students organize information, interpret and solve problems logically. They will also enable the students to come up critical evaluations of the solutions found.

Throughout the chapter, interesting problem contexts serve as the foundation for instruction. As lessons unfold around these problem situations, classroom instruction tends to follow a common pattern as elaborated upon in the instructional procedure.

Classroom activities are designed to actively engage students in problem investigation and making sense of problem situations. They will work together collaboratively in heterogeneous groupings: in pairs or in small groups. They will communicate their mathematical thinking and the results of their group efforts.

Focus Questions 

1. How can polynomial equations be used to provide accurate models of practical problems that involve three dimensions?

2. How can a polynomial model be used to solve problems where maxima or minima are of significant importance?

3. How can polynomial expressions be used to represent and predict social or fiscal changes over time?

Objectives

BEC Standards

After completing Chapter 1, the students should be able to:

➢ identify a polynomial function from a given set of relations;

➢ determine the degree of a given polynomial function;

➢ find the quotient of polynomials by using the division algorithm and synthetic division;

➢ find the quotient using synthetic division and the Remainder Theorem when p(x) is divided by (x-c);

➢ state and illustrate the Remainder Theorem;

➢ find the value of p(x) for x = k by synthetic division and the Remainder Theorem;

➢ state and illustrate the Factor Theorem;

➢ find the zeroes of polynomial functions of degree greater than two by using:

• Factor Theorem

• Factoring

• Synthetic Division

• Depressed Equations

➢ draw the graph of polynomial functions of degree greater than two (use a graphing calculator, if available).

Key Terms

Terms when numbers are added or subtracted, they are called terms. Example: 4x² + 7x − 8 is a sum of three terms.

Factors when numbers are multiplied, they are called factors. Example: (x + 1)(x + 2)(x + 3) is a product of three factors.

Variable is a symbol that takes on values. 

Value is a number; thus if x is the variable and has the value 4, then 5x + 1 has the value 21.

Constant is a symbol that has a single value.

Example:  The symbols '5' and '[pic]' are constants.

The beginning letters of the alphabet a, b, c, etc. are typically used to denote constants, while the letters x, y, z, are typically used to denote variables.  

Example: if we write y = ax² + bx + c, we mean that a, b, c are constants (i.e. fixed numbers), and that x and y are variables.

Monomial in x is a single term of the form axn, where a is a real number and n is a whole number. Examples: 5x3, −6.3x, 2.

Polynomial in x is a sum of monomials in x. Example: 5x3 − 4x² + 7x − 8

Degree of a Term is the sum of the exponents of all the variables in that term. In functions of a single variable, the degree of a term is simply the exponent. Example: The term 5x³ is of degree 3 in the variable x.

Leading Term of a Polynomial is the term of highest degree. Example: The leading term of this polynomial 5x³ − 4x² + 7x − 8 is 5x³.

Leading Coefficient of a Polynomial is the coefficient of the leading term. Example: the leading coefficient of that polynomial is 5.

Degree of a Polynomial is the degree of the leading term. Example: the degree of this polynomial 5x³ − 4x² + 7x − 8   is 3.

Constant Term of a Polynomial is the term of degree 0; it is the term in which the variable does not appear. Example: The constant term of this polynomial 5x³ − 4x² + 7x − 8 is −8.

General Form of a Polynomial shows the terms of all possible degree. Example, is the general form of a polynomial of the third degree: ax³ + bx² + cx + d. Notice that there are four constants: a, b, c, d.

Polynomial function has the form: y = A polynomial

A polynomial function of the first degree, such as

y = 2x + 1, is called a linear function; while a polynomial function of the second degree, such as y = x² + 3x − 2, is called a quadratic.

Domain and range The natural domain of any polynomial function is:

− [pic]< x b, then xa/xb = xa-b

If a < b, then xa/xb = 1/xb - a

RESOURCES

➢ drill board or flash cards

➢ chart

PROCEDURE

Opening Activity

A. To check the students’ understanding of the previous lesson, give some practice exercises on division of polynomials.

Simplify each of the following:

1. 24x7/12x3

2. 12x4y2/4x2y5

3. –42x5y4/6xy3

4. –4(x – y)/20(y – x)4

B. Review the students’ assignment and ask them to solve a new set of exercises on division of polynomials using the long method.

1. (12x2 + 8x – 15) by (x + 3)

2. (x2 + 8x + 16 ) by (x + 4)

3. (x3 + x2 – x – 1) by (x – 1)

4. (2x4 + 5x3 + 11x + 6) by (x + 3)

C. Ask the students to write their solutions on the board using long division. Process the activity by asking these questions:

➢ What is the value of learning the long division method for polynomials?

➢ Would you like to know a shorter way of dividing polynomials?

Main Activity

Present an illustrative example to the class:

Problem: What are the quotient and the remainder when (3x3 + x2 – 8x – 4) is divided by (x – 2)?

Solution

X – 2 [pic]

[pic] [pic]

[pic]

To check, multiply the quotient by the divisor and add the remainder.

(x – 2)(3x2 + 7x + 6) = 3x3 + 7x2 + 6x

-6x2 – 14x – 12

3x3 + x2 – 8x – 12

(3x3 + x2 – 8x – 12) + 8 = = 3x3 + x2 - 8x – 4

we write:

[pic]

Emphasize the fact that since we are dealing with rational expressions, we cannot simply add the remainder, as we do with real numbers. We have to be careful with our notation.

Discussion Ideas

A. Discuss with the students the long division method. Try to explain how to eliminate all variables and write the problem in a more compact form. Let the students compare the long division method used above with the process below. Explain this in detail.

2 3 1 -8 -4

6 14 12

3 7 6 8 or 3x2 + 7x + 6 r8

Show why x – 2 is written as 2 in synthetic division. This is because

x – 2 means x – (2) or x = 2.

B. Present new sets of exercises and ask the students to solve them using synthetic division.

1. Divide (2x3 – 9x2 + 13x – 12) by (x – 3)

Solution:

3 2 -9 13 -12

6 -9 12

2 -3 4 0 or 2x2 – 3x + 4 r 0

2. Divide (x5 – 4x3 + 5x2 – 5) by (x + 1)

Solution:

-1 ( [pic] 1 0 -4 5 0 5

-1 1 3 -8 8

1 -1 -3 8 -8 3

or: x4 – x3 – 3x2 + 8x – 8 r3

3. Find the quotient and the remainder using synthetic division.

6x3 + x2 - 4x + 1 by 3x – 1

Solution

[pic]

Note 3x – 1 = 3[pic]

This means that we can write the above expression as:

|[pic] |2 |[pic] |[pic] |[pic] |

| | |[pic] |[pic] |[pic] |

2 1 -1 0

Thus, [pic]

Check:

(2x2 + x – 1)(3x – 1) = 6x3 + 3x2 – 3x

-2x2 – x + 1

6x3 + x2 – 4x + 1

Guide Questions

1. Which part of the binomial divisor serves as the divisor in this method?

2. What is done with the divisor and the numerical coefficient of the term with the highest degree?

3. Where is the result written?

4. What operations are involved in getting the next numbers?

5. How is the last number obtained? What does a remainder of zero mean?

6. What is the degree of the polynomial quotient?

7. Are there any restrictions in the use of synthetic division? When can we use this method and when is this NOT applicable?

Extension Ideas

Get the meaning of the word “synthetic” from the dictionary. In your own words, write a paragraph explaining why the process discussed earlier is called the “synthetic division process.”

Closing Activity

a) Explain the key learning points in using synthetic division as an alternative method for dividing polynomials.

➢ Synthetic division is a division process for polynomials in one variable when the divisor is of the form x – c, where c is any real number.

➢ Synthetic division is useful in the following cases:

a) finding the quotient and the remainder when a polynomial in x is divided by x – c; and

b) determining if a binomial of the form x – c is a factor of the given polynomial in x.

b) Recalling the steps used in dividing polynomials by synthetic division, ask the students to complete the following:

Steps in Synthetic Division:

1. Write the numerical coefficient of the dividend in ___________order of the exponents with 0 as the coefficient of any missing power of x.

2. Write the___________ of the divisor x – c at the left hand corner of the coefficients.

3. Bring down the ___________of the dividend, multiply it by c and add the result to the second column.

4. Multiply the ______obtained in the previous step by c and add the result to the third column. Repeat this process until you reach the last column of the dividend.

5. The third row of numbers represents the coefficient of the terms in the quotient. The degree of the variable is one less than that of the dividend and the rightmost number is the ________.

Answer Key

1. descending

2. constant term c

3. leading coefficient

4. sum

5. remainder

ASSESSMENT

Find the quotient and the remainder using synthetic division.

a) x3 – 2x2 + 4x + 1 by x – 2

b) 4x3 - 6x2 + 2x + 1 by x – ½

c) 2x4 + 5x3 + 11x + 6 by x + 3

HOMEWORK

Find the quotient and the remainder using synthetic division.

1. x4 + 8x2 – 5x3 – 2 + 15 x by x – 3

2. x5 – 32 by x + 2

3. 2x3 – 7x2 – 8x + 16 by x – 4

4. 3y3 + 2y2 – 32y + 2 by y – 3

5. 76x3 – 19x2 + x + 6 by x – 3

REFERENCES

Foster and Gordon. Algebra 2 With Trigonometry—Applications and Connections

Jose-Dilao. Advanced Algebra, Trigonometry and Statistics. Functional Approach. 72–73.

Yu-hico. Experiencing Mathematics 4.

Lesson 8

THE REMAINDER THEOREM

TIME

One session

SETTING

Math room

OBJECTIVES

At the end of the lesson, the students should be able to:

➢ state the Remainder Theorem; and

➢ use synthetic division and the Remainder Theorem to find the remainder when a polynomial in x is divided by a binomial of the form x – c.

PREREQUISITE

Students should have already learned and applied the steps used in dividing polynomial expressions by means of synthetic and long division.

RESOURCES

➢ Manila paper

➢ marker pen

PROCEDURE

Opening Activity: GUESS-A-REMAINDER

a) Have a drill to check on the students’ basic knowledge on division of whole numbers.

b) Give exercises on how to use synthetic division to find the quotient and the remainder.

1. (2x3 + 5x2 - 4x – 3) by (x + 2)

2. (2a4 + 5a3 + 7a2 – 4a + 6) by (a + 3)

c) Start the new lesson with a simple game following the procedure below:

1. Ask the students to give a 3-digit number.

2. Tell them to divide the number by 9.

Guide Questions

➢ Can you find the remainder without dividing? How?

➢ When will a remainder be equal to zero?

Main Activity: REMAINDER THEOREM

1. Develop the concept of the Remainder Theorem by presenting sample equations. Write and solve these expressions on the board.

a. 48/7 = (6 + r6) ( 48 = 7(6) + r6

b. 17/5 = (3 + r2) ( 17 = 5(3) + r2

Ask the students:

a. Can you write these equations another way?

48 = 7(6) + r6

17 = 5(3) + r2

b. In the new equation, what name is given to each term?

Dividend = (Divisor)(Quotient) + Remainder

2. Allow the students to solve for the quotient and the remainder using synthetic division or long division first. Then, ask them to compare their answers with P(1) in #2.1 and P(-1) in # 2.2

2.1 3x + 5

x – 1

P(x) = 3x + 5

(1) = 3(1) + 5

= 8

Therefore, the remainder is 8.

2.2 2x2 + 4x + 5

x + 1

P(x) = 2x2 + 4x + 5

P(-1) = 2(-1)2 + 4(-1) + 5

= 2 + 1

= 3

Therefore, the remainder is 3.

3. Find the remainder using the Remainder Theorem:

(9a3 + 3a + 6a2 + 9) divided by (3a – 3)

Solution

[pic] (find the remainder)

Can be written as:

[pic]= [pic] (divide the expression by 3 so that the divisor is of the form a – c)

By the Remainder Theorem:

3(1)3 + 1 +2(1)2 + 3 = 3 +1 + 2 + 3 = 9

Multiply this by 3, since 3a – 3 = 3(a – 1)

[pic] The remainder when 9a3 + 3a + 6a2 + 9 is divided by

3a – 3 is 9(3) = 27

Check:

[pic]

- 9a3 – 9a2 _______

15a2 + 3a

-15a2 + 15a___

18a + 9

-18a +18

27

4. Divide the students into small groups. Have them work on the given set of problems provided in the attached Student Activity Sheet.

Discussion Ideas

1. What is the relation between the remainder and the value of the polynomial at x = c when the polynomial in x is divided by a binomial of the form x – c?

2. How is the remainder obtained when the polynomial in x is divided by a binomial of the form x –c?

3. What happens if the remainder is zero?

Note: The value of the polynomial at x = c is equal to the remainder when the divisor is x – c.

Closing Activity

a) Emphasize the key learning points:

➢ Division Algorithm for Polynomials

For each polynomial P(x) of a positive degree, and for any number c, there exist unique polynomials Q(x) and R(x) such that:

P(x) =[ (x – c) (Q(x)] + R(x),

Where: Q(x) is of degree n – 1

R(x) is the remainder

➢ Remainder Theorem

If a polynomial P(x) is divided by x – c, then the remainder is P(c).

Proof of the Remainder Theorem

1. P(x) = (x – c) ( Q(x) + R(x) Division Algorithm for Polynomials

2. P(x) = (x – c) ( Q(x) + R Definition of Division of Polynomials

R(x) must have a degree less than the

degree of (x – c).

` Thus, R(x) = R is a constant (which may or may not be 0)

3. P(c) = [(c – c) ( Q( c )] + R Equation (2) is true for all x. Therefore x = c

4. P(c) = 0 x Q ( c ) + R Multiplication Property

5. P(c) = R Identity

Hence, the remainder R is equal to P(c).

b) Ask the students to relate synthetic division with the Remainder Theorem in terms of evaluating polynomials.

Answer Key

➢ Synthetic division can be used as a convenient way to find the values of polynomial functions P(c). Hand in hand with this process is the Remainder Theorem. The remainder r obtained in synthetic division is indeed equal to P(c).

ASSESSMENT

Find the remainder using the Remainder Theorem.

1. (2y3 – 5y2 – 8y – 50) divided by (y – 5)

2. (3y3 + 2y2 – y + 5) divided by (y + 2)

3. (x3 + 4x – 7) divided by (x – 3)

4. (8x4 + 2x + 4x3 + 1) divided by (x + ½)

HOMEWORK

Using the Remainder Theorem, find the remainder of the following:

1. (12x4 + 11x – x2 – 40x3 –90) by (x – 3)

2. (2x4 – 4x3 + 9x2 + 2x – 5) by (x – 1)

3. (4y3 – 6y2 + 2y + 1) by (2y – 1)

4. (y4 + 8y2 – 5y3 – 2 + 15y) by (y – 3)

5. (3m3 – 2m2 + 2m – 3) by (m + 3)

REFERENCES

Foster, Gordon. Algebra 2 With Trigonometry—Applications and Connections.

Jose-Dilao. Advanced Algebra, Trigonometry and Statistics. Functional Approach. 74–77.

Yu-hico. Experiencing Mathematics 4.

Student Activity 8

REMAINDER THEOREM APPLICATIONS

Solve the following polynomial functions:

A. Use the Remainder Theorem to find the value of x4 – 4x2 + 12x – 9

when x =2.

B. Use the Remainder Theorem to find the value of x4 – 4x3 + 12x – 9

when x = 1.

C. Determine whether x – 3 is a factor of 2x3 – 3x2 – 12x + 9.

If x – 3 is a factor, name the other factor of the polynomial.

D. Which of the following binomials are factors of x4 – 3x2 + 6x - 4?

a. x + 2

b. x – 1

c. x + 3

Lesson 9

THE VALUES OF POLYNOMIAL FUNCTIONS

TIME

One session

SETTING

This lesson must be done inside the math room. This is simply a continuation of the past two lessons on synthetic division and the Remainder Theorem. If the class learned the past lessons quickly, this lesson may be integrated into those lessons as no new concepts are presented here.

OBJECTIVES

At the end of the lesson, the students should be able to further develop their skills in determining the:

➢ quotient of a polynomial divided by a binomial of the form (x - c) using synthetic division; and

➢ value of a polynomial using synthetic division and the Remainder Theorem.

PREREQUISITE

Students should have already learned the Remainder Theorem and synthetic division.

RESOURCES

➢ Manila paper

➢ marker pen

PROCEDURE

Opening Activity

A. Check the students’ assignments and review the class on evaluating polynomial functions by means of a short contest:

B. Form two teams to evaluate polynomials for integral values of the variable. One team will use direct substitution while the other team will use synthetic division. Ask the students to work on the following practice set.

Task: Determine the remainder in each of the following polynomial expressions using synthetic division.

a. y3 – 4y2 – 2y + 5; y – 1

b. 2x3 – 3x2 + 2x – 8; x – 2

c. 4m3 + 4m2 + m + 3; m – 4

Main Activity

A. Give the following set of questions for students to solve as a group. The outputs of each group will be presented to the class.

1. Given the polynomial 8x3 + 6x + 7, find its degree and leading coefficient.

(Answer: Since 3 is the highest exponent, the polynomial is of degree 3 and 8 is its leading coefficient.)

2. To find the quotient and the remainder when 3x3 – 5x + 10 is divided by x – 2, what are the initial steps you must take?

(Answer: Arrange the terms in order of descending power and write zero for the missing terms.)

3. Using long division and synthetic division, find the quotient and remainder. Compare your answers.

Long Division

x – 2 [pic]

[pic]

[pic]

[pic]

Synthetic Division

2 3 0 - 5 10

6 12 14

3 6 7 24

3x2 + 6x + 7 24 is the remainder

3x2 + 6x + 7 is the quotient

[pic] The quotient is 3x2 + 6x + 7 and the remainder is 24.

Using either of the methods, students should be able to find the same quotient and the remainder.

5. Using the Remainder Theorem, find the value of the polynomial at x = 2.

f(x) = 3x3 – 5x + 10

f(2) = 3(2)3 – 5(2) + 10

= 3(8) – 10 + 10

f(2) = 24

By the Remainder Theorem, the remainder will be equal to the value of the polynomial at x = 2.

B. Present another example:

Divide (2x3 – 4x2 – 5x + 5) by (x + 1)

Since synthetic division is much shorter, you may use this method to save time.

-1 2 -4 -5 5

-2 6 -1

2 -6 1 4

2x2 – 6x + 1 r 4

f(x) = 2x3 – 4x2 – 5x + 5

f(-1) = 2(-1)3 – 4(1)2 – 5(-1) + 5

= 2(-1) – 4(1) + 5 + 5

= –2 –4 + 5 + 5

= –6 + 10

f(-1) = 4

Discussion Ideas

1. When we divided 3x3 – 5x + 10 by x – 2, what did we use as divisor in the synthetic division?

2. What can you say about the remainder and the value of the polynomial?

Closing Activity

1. Synthesize the key points of the lesson by saying:

If a polynomial f(x) is divided by x – r until a remainder without x is obtained, this remainder is equal to f (r).

2. Ask students: How do you find the value of a polynomial?

Answer: Apply the Remainder Theorem and synthetic division.

ASSESSMENT

Find the value of the polynomials using the Remainder Theorem and synthetic division.

1. (2y3 – 5y2 – 8y – 50) divided by (y – 5)

2. (3y3 + 2y2 – y + 5) divided by (y + 2)

3. (x3 + 4x – 7) divided by (x – 3)

4. (9a3 + 3a + 6a2 + 9) divided by (3a – 3)

5. (8x4 + 2x + 4x3 + 1) divided by (x + ½)

HOMEWORK

Using the Remainder Theorem, find the value of the polynomials.

1. (12x4 + 11x – x2 – 40x3 –90) by (x – 3)

2. (2x4 – 4x3 + 9x2 + 2x – 5) by (x – 1)

3. (4y3 – 6y2 + 2y + 1) by (2y – 1)

4. (y4 + 8y2 – 5y3 – 2 + 15y) by (y – 3)

5. (3m3 – 2m2 + 2m – 3) by (m + 3)

REFERENCES

Foster, Gordon. Algebra 2 With Trigonometry–Applications and Connections

Jose-Dilao. Advanced Algebra, Trigonometry and Statistics. Functional Approach. 74–77.

Yu-hico. Experiencing Mathematics 4

Lesson 10

THE FACTOR THEOREM

TIME

One session

SETTING

Math room

OBJECTIVE

At the end of this lesson, the students should be able to use the Factor Theorem and other concepts to find the zeroes of the polynomial function of degree greater than two.

PREREQUISITE

Students should have learned the zeroes of polynomial functions.

RESOURCES

➢ Chart

PROCEDURE

Opening Activity

A. Have students practice their skills in factoring by solving the following:

➢ 4y – 24

➢ 15a3 + 20a

➢ 4x2 –81

➢ y2 + 18y+ 81

B. Conduct a review on finding the zeroes of a polynomial function. Present these examples as written on Manila paper for the students’ verification.

1. Given f(x) = 4x + 2, find the zero of the function.

4x + 2 = 0

4x = 2

x = -2/4 or –1/2

1. If f(x) = x2 + 3x – 10, find the zeroes of the function:

x2 + 3x – 10 = 0

(x + 5)(x – 2) = 0

x = –5; x = 2

The zeroes of the function are –5 and 2.

Main Activity

A. Begin the new lesson by saying:

A linear function has only one zero, while a quadratic function, has at most two zeroes. If the degree of a given polynomial function is n, how many zeroes does the function have? How are the zeroes of a function found?

a. What is/are the zero/zeroes of the function in the previous examples?

b. How many zeroes are there in each function?

B. Develop the students’ understanding of the Factor Theorem by asking them to analyze the example below (already prepared in Manila paper for the presentation).

a) P(x) = x3 – x2 – 10x – 8

b) P(x) = x3 + x2 – x – 1

C. Discuss the Factor Theorem and relate this to the Remainder Theorem. Explain how these two theorems complement each other. Also discuss how they can be used to solve problems involving roots and the factorization of polynomials with degree higher than 2.

The Remainder Theorem states that:

“When the polynomial P(x) is divided by x-c, the Remainder is P(c)”

The Factor Theorem

“Given the polynomial P(x). If P(c) = 0, then x-c is a factor of P(x). Conversely, if x-c is a factor of P(x), then P(x) = 0.

If the remainder when P(x) is divided by (x-c) is zero, then (x-c) is a factor of P(x).

PROOF

Suppose P(c) = 0. By the Remainder Theorem, when P(x) is divided by (x–c), the remainder R = P(c) = 0

Then, P(x) = (x-c) ( Q (x) + R becomes

P(x) = (x–c) ( Q (x) + 0

P(x) = (x–c) ( Q (x)

Therefore (x–c) is a factor of P(x)

D. Present some illustrative examples depicting the Factor Theorem

1) Ask the students to guess the value of x in the given function: P(x) = x3 – x2 – 10x – 8

Substitute 4, -2, -1 to the given function.

If x = 4,

x3 – x2 – 10x – 8

43 – 42–10(4) – 8 = 0

(x – 4) is a factor of P(x).

If x = -2,

X3 – x2 –10x – 8

(-2)3 – (-2)2 – 10(-2) – 8 = 0

(x + 2) is a factor of P(x)

If x = -1

x3 – x2 – 10x – 8

(-1)3 – (-1)2 – 10(-1) – 8 = 0

(x + 1) is also a factor of P(x).

Therefore, the zeroes are 4, –2, and –1.

2) P(x) = x3 + x2 – x – 1

Substitute 1, 2, -1 to the given function.

If x = 1

x3 – x2 – x – 1

13 + (1)2 – 1 – 1

1 + 1 – 1 – 1 = 0

(x–1) is a factor of P(x)

If x = 2

x3 + x2 – x – 1

23 + 22 – 2 – 1

8 + 4 – 2 – 1

12 – 2 – 1 = 9

(x – 2) is not a factor of P(x) and 2 is not a zero of P(x)

If ax = -1

x3 + x2 – x – 1

(-1)3 + (-1)2 – (-1) – 1

-1 + 1 + 1 –1 = 0

(x + 1) is a factor of P(x).

➢ Therefore, the only zeroes of the function are 1 and –1.

➢ f(x)=x3 + x2 – x – 1 is a polynomial of degree 3.

➢ We found that 1 and –1 are roots of this equation.

➢ Using synthetic division, we have:

|1 |1 |1 |-1 |-1 |

| | |1 |2 |1 |

|-1 |1 |2 |1 |0 |

| | |-1 |-1 | |

|-1 |1 |1 |0 | |

| | |-1 | | |

1. 0

Note: From this, we can see that p(x) can be factored as:

➢ p(x) = x3 + x2 – x – 1 = (x – 1)(x +1)(x + 1)

➢ because –1 served as a root twice, we say –1 is a root of MULTIPLICITY 2

E. Divide the class into small groups. Have the groups work on the following sets of polynomial functions:

➢ Which of the following binomials, (x – 1), (x – 2), (x + 2), (x –3), (x + 3) are divisors of the following expressions? Find the zeroes of their related functions.

a. x3 + 2x2 – 9x – 8

b. x4 – 4x3 – x2 + 16x – 12

c. x4 – 2x3 – 7x2 + 8x + 12

Discussion Ideas

1. What are the degrees of the functions in numbers 1 and 2?

2. How many zeroes are there in numbers 1and 2?

3. How are 4, -1, -2 related to the constant term given in example number 1? How are 1, 2, -1 related to the constant term in example number 2?

4. Why is there a need to substitute these possible factors to the function?

5. How are the zeroes of a polynomial function related to the constant term of the polynomial?

6. When is a certain number a zero of a polynomial function? When is it not a zero?

7. What are the steps in finding the zeroes of P(x) using the Factor Theorem?

Closing Activity

Ask students to determine the steps in finding the zeroes of a polynomial P(x). Consolidate their responses to come up with the following steps:

1. Substitute possible values of c for x in the given polynomial P(x).

1. Use the Remainder Theorem. If P(c) = 0, then (x – c) is a factor of P(x) by virtue of the Factor Theorem.

2. Equate each factor to zero to get the zeroes of the function.

3. The number of zeroes of a function is less than or equal to the degree of the given function.

ASSESSMENT

Find the zeroes of the function:

1. P(x) = x3 + 3x2 – 2x – 6

2. P(x) = x3 – x2 – 4x + 4

3. P(x) = 3x3 – x2 + 12 – 4

HOMEWORK

Find the zeroes of the following polynomials using the Factor Theorem.

1. P(x) = x3 – 12x2 + 41x - 4

2. P(x) = x4 – 3x2 + 2

REFERENCES

Foster, Gordon. Algebra 2 With Trigonometry–Applications and Connections

Jose-Dilao. Advanced Algebra, Trigonometry and Statistics. Functional Approach. 82–84.

Mathematics IV. Advanced Algebra, Trigonometry and Statistics. 2002 BEC.

Yu-hico. Experiencing Mathematics 4.

Lesson 11

ZEROES OF POLYNOMIAL FUNCTIONS: A RECALL

TIME

One session

SETTING

Math room

OBJECTIVES

At the end of this lesson, the students should be able to:

➢ apply the Factor Theorem, factoring techniques, synthetic and long division and the Rational Root Theorem; and

➢ find the zeroes of polynomial functions of degree greater than 2.

PREREQUISITE

The students are expected to have learned the following concepts:

➢ Finding the zeroes of linear and quadratic functions, synthetic and long division as well as the Remainder Theorem to solve for the quotient and the remainder of a polynomial function

RESOURCES

➢ graph board

➢ chart

➢ colored chalk

PROCEDURE

Opening Activity

A. Check the students’ assignment and review finding the quotient and the remainder, if necessary. Ask the students to find the quotient and the remainder of the following polynomials:

a. (2y3 – 5y2 – 8y – 50) divided by (y – 5)

b. (3y3 + 2y2 – 7) divided by (y – 3)

B. Introduce the lesson by reviewing how to find the zeroes of a polynomial function.

➢ Recall the meaning of “zeroes of a polynomial function”. Ask the students for illustrations of quadratic and linear functions.

➢ Ask them to extend the meaning of zeroes of polynomials of degree > 2 by illustrating this graphically.

➢ Present the graphs of the following linear and quadratic functions, and ask the students to answer the guide questions that follow.

Graphs of Quadratic Functions

Guide Questions

For each function:

➢ At what point/points does the graph cross the x-axis?

➢ Notice that in each of the points of intersection with the y-axis the value of f(x) is zero. What is the value of x when f(x) = 0?

➢ What are these x-values called?

1. Ask the students to find the value of x when f(x) = 0

a. Find the zero of f(x) = 3x – 5.

F(x) = 3x – 5

0 = 3x - 5

3x = 5

x = 5/3

a. Find the zeroes of f(x) = x2 – 5x + 4

F(x) = x2 – 5x + 4

x2 – 5x + 4 = 0

(x – 4)(x – 1) = 0

x = 4, x = 1

( The zeroes of the function are 4 and 1

1. Illustrate the procedure for finding the zeroes of a polynomial function by factoring. Let the students find the zeroes of the following functions:

a) P(x) = (x – 5)2 (x + 3) Zeroes are 5 and –3, 5 being a double root

b) P(x) = 3x3 + 9x2 – 30x Zeroes are 0, -5 and 2.

2. Recall the Factor Theorem, which states that:

“Given the polynomial f(x). If the remainder when f(x) is divided by (x-c) is zero, then (x-c) is a factor of f(x).

Main Activity

Part I

Let individual students try out some integers in the synthetic division for each polynomial and find the zeroes of the given P(x):

a. P(x) = x3 + 9x2 + 23 x + 15

b. P(x) = x4 + 5x3 + 5x2 – 5x – 6

Answers

a. Zeroes are –1, -3, and –5

b. Zeroes are 1, -1, -3, and –2.

Guide Questions

➢ For P(x) = x3 + 9x2 + 23x + 15, what numbers may possibly be eliminated? (0 and positive values). Why?

➢ How many negative integers should one try out as possible zeroes?

➢ When should one stop dividing synthetically?

Discussion Ideas

Based on the previous lesson, the students were able to find the factors of polynomials of higher degree. At this point, the students shall find the zeroes of such polynomials.

➢ The polynomial f(x) = 4x3 – 4x2 – 8x is of the third degree; therefore, we expect the function to have three zeroes.

➢ Look for a common monomial factor among the terms of the function 4x3 – 4x2 – 8x and factor this out: 4x (x2 - x – 2).

➢ Note that x2 – x – 2 can be factored further as (x – 2)(x + 1).

Therefore, 4x3 – 4x2 – 8x = 4x (x – 2)(x + 1)

➢ Equating all the factors to zero since f(x) = 0 we find that if:

4x = 0 x – 2 = 0 x + 1 = 0

Then:

x = 0 x = 2 x = -1

Note: This method is applicable if the given polynomial can easily be factored.

Part II. Present another example for the students.

f(x) = x4 – x3 – 7x2 + x + 6

The factors of 6 are: 2 and 3, -2 and -3, 6 and 1, -6 and -1?

Solution

Because f(x) is not readily factorable, we make use of a theorem called the RATIONAL ROOT THEOREM:

Rational Root Theorem

Given a polynomial p(x0 = anxn+an-1xn-1 + … + a0. Then the rational roots of the polynomial are of the form [pic] where bo and bn are divisors of ao and an, respectively.

We apply this theorem to our examples:

➢ fx) = x4 – x3 – 7x2 + x + 6 where, an = 1 and a0 = 6.

➢ The divisors of ao = 6 are: 1, 2, 3, 6, -1, -2, -3, -6

➢ The divisors of ao = 1 are: 1, -1

The only possible rational roots of f(x) are: [pic]1, [pic]2, [pic]3, [pic]6

We choose which of these numbers to use in synthetic division:

Note

1. Clearly, 1 is a root, since f(1) = 1 – 1 – 7 + 1 + 6 = 0. Thus, the remainder when f(x) is divided by (x-1) is zero. By the FACTOR THEOREM, (x – 1) is a factor of f(x).

1. Similarly, -1 is a root since f(-1) = 1 + 1 – 7 – 1 + 6 = 0

Because 1 is a root, we look for the quotient [pic] by Synthetic Division

|1 |1 |-1 |-7 |1 |6 |

| | |1 |0 |-7 |-6 |

|-1 |1 |0 |-7 |-6 |0 |

| | |-1 |1 |6 | |

| |1 |-1 |-6 |0 | |

Note: We can continue the synthetic division in this way until all the rational roots have been found.

So far, we found two roots of f(x): 1 and –1.

Using the Factor Theorem, we know that:

f(x) = (x – 1)(x + 1) (x2 – x – 6)

= (x – 1)(x + 1) (x –3)(x + 2)

Thus, the roots of f(x) are 1, -1, 3 and –2.

Note: After the above example, the teacher may give another similar example for the students to work on by themselves.

Example

Find all the zeroes of f(x) = x4 + 2x3 – 3x2 – 8x – 4. Use all available techniques.

Solution

a) By the Rational Root Theorem, an = 1, ao = –4

( possible rational roots [pic]1, [pic]2, [pic]4.

b) By using the Remainder Theorem: we try to guess the roots

f(1) = 1 + 2 – 3 – 8 – 4 = 3 – 3 – 8 – 4 ( 0 ( 1 is not a root

Thus, x – 1 is NOT a factor of f(x))

f(-1) = 1 - 2 – 3 + 8 – 4 = 1 – 5 + 8 – 4 = 0

[pic] -1 is a root of f(x), and so (x + 1) is a factor of f(x)

c) Using Synthetic Division find the quotient:

|-1 |1 |2 |-3 |-8 |-4 |

| | |-1 |-1 |4 |4 |

|-1 |1 |1 |-4 |-4 |0 |

| | |-1 |0 |4 | |

| |1 |0 |-4 |0 | |

Note: It is possible that a number is a root more than once. In this case, -1 is of MULTIPLICITY 2 because it is a root 2 times

So far, we know by Synthetic Division and the Factor Theorem that:

f(x) = (x + 1)(x + 1)(x2 – 4) = (x + 1)(x + 1)(x + 2)(x – 2)

Hence, the roots of f(x) = x4 + 2x3 – 3x2 – 8x – 4 are –1, –2 and 2, where –1 is of multiplicity 2.

Part III. Divide the students into small groups. Ask them to find the zeroes of each of the given functions:

1. f( x ) = 2x3 – x2 – 13x – 6

2. f( x ) = 4x3 + 13x2 – 37x – 10

3. f( x ) = 6x3 + 25x2 + 2x – 8

4. f (x) = 6x3 + 19x2 + 11x - 6

Guide Questions

➢ What are the ways of determining the zeroes of a function?

➢ Without graphing, how are the zeroes of a function determined?

➢ How many zeroes are there in a:

• linear function?

• quadratic function?

• polynomial of degree n?

Closing Activity

Ask the students: What is meant by “zeroes of a function”?

1. A zero of a polynomial function is the value of the variable x, which makes the polynomial equal to zero, or f(x) = 0.

2. The zeroes of P(x) are the roots of the equation P(x) = 0. A linear function has only one zero, while a quadratic function has at most two zeroes.

3. The zeroes of quadratic function can be found by factoring, completing the square, or by using the quadratic formula.

4. In general, a polynomial of degree n will have at most n real roots or zeroes.

ASSESSMENT

A. Factor the following polynomials into linear factors:

1. f(x) =x4 + 3x3 – 3x2 – 7x + 6

1. f(x) =x4 + x3 – 11x2 – 9x + 18

B. Find the zeroes of the following:

a. f(x) = x + 3

b. f(x) = x2 – x – 30

c. f(x) = x2 – 2

d. f(x) = x2 + 4x + 4

HOMEWORK

Graph the following functions:

1. f(x) = 2x – 7

2. f(x) = 3x + 8

3. f(x) = x2 – 8x – 48

4. f(x) = x2 + 9x - 22

REFERENCES

Foster, Gordon. Algebra 2 With Trigonometry–Applications and Connection

Jose-Dilao. Advanced Algebra, Trigonometry and Statistics. Functional Approach. 78–87.

Yu-hico. Experiencing Mathematics 4.

Lesson 12

GRAPHS OF POLYNOMIAL FUNCTIONS (Part 1)

TIME

One session

SETTING

Math room

OBJECTIVES

At the end of the lesson, the students should be able to:

➢ graph the polynomial functions of degree greater than two; and

➢ locate the zeroes of a polynomial function by estimating its position on a graph

PREREQUISITE

The students must be familiar with the Factor Theorem, the Remainder Theorem, the Rational Root Theorem, and with synthetic and long division

RESOURCES

➢ chart

PROCEDURE

Opening Activity

a) Check the students’ basic knowledge on evaluating a polynomial function. Ask them to find the value of P(x):

Given: P(x) = x3 – 3x2 – x + 10

Find: P(2), P(-2), P(3)

b) Introduce the new lesson by asking students to illustrate the graphs of the following:

a) F(x) = a?

b) F(x) = ax + b?

c) F(x) = ax2 + bx + c?

d) F(x) = x3?

c) Allow the students to illustrate the different cases or possibilities for the first three. Let the students graph each of the functions.

Main Activity

Note! Prepare sample graphs for presentation to the students:

1. Point out the zeroes of the functions graphed on the board. Then, explain the zeroes of the polynomial functions.

2. Let the students observe the graphs of the polynomial function

f(x) = xn, when:

a) n is an integer greater than zero.

b) n is even

c) n is odd

3. Give some examples:

a. Consider the function f(x) = x4 – 5x2 + 4

Using the different techniques for finding the roots of a polynomial function, the roots of f(x) are found to be: –2, -1, 1 and 2.

The graph of f(x) looks like this:

[pic]

Notice that the graph of f(x) intersects the x-axis at the zeroes of the function (f(x)= x4 – 5x2 + 4) or where y is equal to zero. Because f(x) is of degree 4, it has, at most, 4 real roots.

b. On the other hand, the graph of f(x) = x4 – 2x3 – 2x – 1 looks like this:

➢ How many zeroes does f(x) have? Try to verify your answer by using previous techniques learned.

4. Individual Activity

a) Ask the students to follow the steps in graphing the polynomial function: f(x) = (x – 2) (x + 1)(x – 1), as described in the textbook.

Remind them that the zeroes of a polynomial function are the roots of the related polynomial equation.

b) Find the zeroes of the following polynomial functions, then graph.

➢ P(x) = x3 – 9x + 1

➢ P(x) = x3 – 4x - 2

Discussion Ideas

1. Describe the graph of a polynomial function of degree greater than 2. Differentiate it from linear and quadratic functions.

2. What is meant by “the zeroes of a function”? How can you determine the zeroes from the graphs?

3. What does the graph of f(x) = xn look like when n is even? What if n is odd?

4. How would you describe the graph of f(x) = x5? What about

f(x) = x6?

5. What is the basic step in graphing any function?

Closing Activity

Ask the students:

What can you say about the graphs of polynomial functions?

➢ The graph of a polynomial function has the equation of the form f(x)=x n, where n is an integer greater than zero. It is continuous such that it has no breaks and has smooth rounded turns.

ASSESSMENT

Graph the function P(x) = x3 – 3x + 1, and estimate the real roots to the nearest one-half unit.

HOMEWORK

Use a graph to estimate the real roots to the nearest half unit of the equation P(x) = 0.

➢ P(x) = x3 – x2 – 2x – 1

➢ P(x) = x3 – x2 – 2x + 1

REFERENCES

Foster and Gordon. Algebra 2 With Trigonometry–Applications and Connections.

Jose-Dilao. Advanced Algebra, Trigonometry and Statistics. Functional Approach. 88–92.

Math IV: Advanced Algebra, Trigonometry and Statistics–2002 BEC

Yu-hico. Experiencing Mathematics 4.

Lesson 13

GRAPHS OF POLYNOMIAL FUNCTIONS (Part 2)

TIME

One session

SETTING

This lesson can be done inside the math room. This lesson is supplementary to the previous lesson. The teacher should focus more on polynomial functions of degree >2 and not spend too much time recalling quadratic functions.

OBJECTIVES

At the end of this lesson, the students should be able to:

➢ sketch the graph of a given polynomial function by assigning points;

➢ apply previously learned theorems and concepts in graphing polynomials of degree > 2.

PREREQUISITE

The students must have enough skills in applying the Factor Theorem, factoring techniques, synthetic division and depressed equations to find the zeroes of polynomial functions of degree greater than two.

RESOURCES

➢ graphing boards

➢ graphing papers

➢ colored chalk

➢ chart

PROCEDURE

Opening Activity

A. As a practice exercise, ask the students: Which of the following functions is linear? Which are quadratic?

1. P(x) = 3x + 5

2. P(x) = x2 – 5x + 4

3. P(x) = x2 - 4

B. To review the graph of a polynomial function, show 5 samples of graphs of linear and quadratic equations. Let the students identify the kinds of graphs presented.

1) Present three linear functions, e.g.:

➢ f(x) = x + 2

➢ f(x) = -2x – 1

➢ f(x) = x – 5

Then ask the students to draw the graphs of each of these and let them identify the following:

a. line

b. slope

c. x-intercept

d. y-intercept

Ask the students: How is the graph of a linear function drawn?

1) Present two quadratic functions, such as:

➢ f(x) = x2 – 3x – 4

➢ f(x) = -x2 + 2x + 24.

Give the students a few minutes to draw their graphs and let them identify the following:

a) parabola

b) vertex

c) axis of symmetry

d) direction of opening

e) x-intercepts

Ask the students: How is the graph of a quadratic function drawn?

C Present a third-degree polynomial function in factored form, e.g., P(x) = (x – 2)(x +1)(x – 1). Prepare a Cartesian coordinate plane on the board and guide the students in sketching the graph of this function.

Main Activity

A. Compare the graphs of linear and quadratic functions:

➢ From previous discussions, we learned of the graphs of two polynomial functions, the linear and the quadratic functions. The graph of a linear function f(x) = ax + b, where a and b are real numbers, is a straight line; the graph of a quadratic function, f(x) = ax2 + bx + c, where a, b, and c are real numbers and a is not zero, is a parabola

B. Allow each group to work on the following problems:

1. What do you think will happen if:

a) g(x) is decreased by 2 as in g(x)= x3 – 2?

b) h(x) is increased by 3 as in h(x) = x3 + 1

1. Sketch each of them and compare them with the graph of

f(x) = x3.

Discussion Ideas

a. Consider the following important points:

1) The zeros of a function are really the x-intercepts of the function. These can be plotted easily.

2) The values of the function in the following intervals will provide a picture of the behavior of the graph in these intervals.

x < -1

-1< x < 1

1 < x < 2

x > 2

b. Have the students prepare a table of values where 2 to 3 values for x from each of the above intervals are used. These should be enough for the students to have an idea of the behavior of the graph in each interval. Let them show the table of values and the resulting graph:

Closing Activity

a. Ask the students: How do you graph a given polynomial function?

b. Summarize their responses:

➢ To graph a polynomial function, first find the zeroes of the function. Then construct a table of values of the given variables and the corresponding value of the polynomial function. These values should represent points in the different intervals into which the zeroes of the function divide the x-axis. Plot these values on a Cartesian coordinate plane. If the points are rather far apart, assign fractional values to the variable to determine the shape of the curve more accurately.

ASSESSMENT

1. Draw the graph of the polynomial function: f(x) = x3- 4x2 + x + 3

2. What are the roots of the equation?

3. Show the table of values.

4. What are the x-intercepts?

HOMEWORK

1. Graph the function f(x) = [pic]

2. Graph the function f(x) = x3 + 2x2 – 4x – 1

3. Graph the function f(x) = (x – 1)4

REFERENCES

Foster and Gordon. Algebra 2 With Trigonometry—Applications and Connections

Jose-Dilao. Advanced Algebra, Trigonometry and Statistics. Functional Approach. 88–92.

Math IV: Advanced Algebra, Trigonometry and Statistics – 2002 BEC

Yu-hico. Experiencing Mathematics 4.

Lesson 14

GRAPHS OF POLYNOMIAL FUNCTIONS (Part 3)

TIME

One session

SETTING

Math room

OBJECTIVE

At the end of this lesson, the students should be able to:

➢ solve polynomial functions using the most logical procedure; and

➢ graph polynomial functions of degree greater than two.

PREREQUISITE

Students must have enough skills in applying the Factor Theorem, factoring techniques, synthetic division and depressed equations to find the zeroes of polynomial functions of degree greater than two.

RESOURCES

➢ graphing boards

➢ graphing paper

➢ colored chalk

➢ chart

PROCEDURE

Opening Activity

A. Have students review plotting of points. Ask them to identify the following features of the graph of a polynomial function:

a. vertex

b. axis of symmetry

c. x-intercepts

d. y-intercepts

B. The teacher may choose to solve items in the assignment.

Main Activity

Students have learned to identify the zeroes and turning points of a function from a prepared graph. With this, they can proceed to the next activity. Present the solution and graph of a polynomial function to the students for observation. Example: f(x) = x3 – 3x + 2

We can use the Rational Root Theorem to find out the possible roots of f(x). In f(x) = x3 –3x+ 2, the factors of 2 are 1, -1, 2, -2. We test each of these to determine the roots:

f(x) = x3 – 3x + 2

f(1) = 13 – 3(1) + 2

= 1 – 3 + 2

= 0, then 1 is a zero

f(x) = x3 – 3x + 2

f(-1)= (-1)3 – 3(-1) + 2

= -1 + 3 + 2

= 4, then –1 is not a zero

f(x) = x3 – 3x + 2

f(2) = 23 – 3(2) + 2

= 8 – 6 + 2

= 4, then 2 is not a zero

f(x) = x3 – 3x + 2

f(-2) = (-2)3 – 3(-2) + 2

= -8 + 6 + 2

= 0, then –2 is a zero

Ask the students to prepare a table of values using numbers representing the various intervals. Then, plot the graph of the function.

Discussion Ideas

Ask the students to look at the graph and to answer the following questions:

1) What are the zeroes of the function?

2) What happens to the graph in each interval?

3) What are the factors of f(x)?

4) Which root has a multiplicity of 2?

Extension Ideas

Ask the students to look for a problem in science that make use of the zeroes of a polynomial function.

Closing Activity

Emphasize the key points of the lesson by asking:

1) How do you graph a given polynomial function?

2) How do you find the zeroes of a polynomial function?

ASSESSMENT

Solve and graph the function f(x) = x3 + 6x2 + x + 6

➢ What are the zeroes of the function?

HOMEWORK

Solve and graph:

1. f(x) = x3 – 8

2. f(x) = x4 + 3x3 + 2x2 - 2

REFERENCES

Foster and Gordon. Algebra 2 With Trigonometry—Applications and Connections,

Jose-Dilao. Advanced Algebra, Trigonometry and Statistics. Functional Approach. 88–92.

Math IV: Advanced Algebra, Trigonometry and Statistics – 2002 BEC.

Mathematics IV – SEDP Series.

Yu- hico. Experiencing Mathematics 4.

Lesson 15

GRAPHS OF POLYNOMIAL FUNCTIONS (Part 4)

TIME

One session

SETTING

Math room

OBJECTIVES

At the end of this lesson, the students should be able to accurately draw the graph polynomial functions of degree greater than two.

PREREQUISITE

Students must have enough skills in applying the Factor Theorem, factoring techniques, synthetic division and depressed equations to find the zeroes of polynomial functions of degree greater than two.

RESOURCES

➢ Manila paper

➢ graphing papers

➢ pencil

➢ ruler

PROCEDURE

Opening Activity

1. Check the students’ assignments.

2. Divide the class into 5. Let each group draw the graph of any given polynomial function on A3–sized bond paper. Let them emphasize the curve using a marker pen or crayon. Ask them to create an artwork using the curve as background.

3. Ask each group to describe its graph. They may add something like:

➢ In graphing polynomial functions, the zeroes of the polynomial function are the roots of the polynomial equation which corresponds to f(x). The real roots of any polynomial equation correspond to the point of intersection of its graph and the x-axis.

Main Activity

Group the students into 10. Give them 30 minutes to work on the given polynomials. Assign a leader in each group and a reporter to discuss the output of the group.

Tasks

1. Graph the polynomial function: f(x) = 4x3 + 16x2 + 9x – 9

2. Sketch the graph of the polynomial function: f(x) = x3 + 3x2 - x – 4

Closing Activity

a) Emphasize the key points of the lesson by asking:

1) How do you graph a given polynomial function?

2) How do you find the zeroes of a polynomial function?

b) Summarize their responses as follows:

➢ To graph a polynomial function, first find the zeroes of the function. Then construct a table of values of the given variable and the polynomial function. These values should represent points in the different intervals into which the zeroes of the function divide the x-axis. Plot these values on a Cartesian coordinate plane. If the points are rather far apart, assign fractional values to the variable to determine the shape of the curve more accurately.

HOMEWORK

1. Graph the function y = x2 + 3x2 –x –3 using (-3,-2,0,1,2,3) as the domain.

2. Approximate the real zeroes by graphing the polynomial function:

y = x3 – 4 x2 + x + 5

REFERENCES

Foster and Gordon. Algebra 2 With Trigonometry—Applications and Connections

Jose-Dilao. Advanced Algebra, Trigonometry and Statistics. Functional Approach. 91–92.

Math IV: Advanced Algebra, Trigonometry and Statistics – 2002. BEC.

Mathematics IV – SEDP Series.

Yu-hico. Experiencing Mathematics.

Unit Integration Plan 

Objectives

While studying polynomials, students can:

➢ study drug absorption rates;

➢ investigate how volcanic eruptions are predicted;

➢ apply polynomials to situations involving population growth;

➢ explore, analyze, and compare the benefits and liabilities of annuities;

➢ analyze, solve and perform real-world problems.

Procedure

Ask each group to solve the following problems:

1. Given the prospective earnings and the prevailing interest rate, determine the amount that will be available for graduate school expenses if a college student saves income from summer employment following high school, until he/she enters graduate school after four years of college.

2. Given the sum of the length, width, and depth of carry-on luggage, determine the real domain for these variables as well as the maximum volume of a carry-on item if the length is to be 10 centimeters more than the depth.

3. Plot the given data on the number of minutes needed to have Caucasian skin tanned at different times of the day in Boracay. Use a graphing calculator to determine a polynomial function to graph the data. Use the function to determine the time of day when skin will tan in 30 minutes. Explain why the time would be the same or different in different parts of the Philippines.

 

4. Given three patterns for constructing closed rectangular containers from pieces of cardboard, determine which of the three patterns produces a container with maximum volume and minimum waste.

Using the computed values, students can be asked to apply their mathematical skills in making bags, ornamental boxes and useful containers which they can sell. They can use any available resources to produce these products and their creativity in making the best appropriate design. Accuracy and quality of work are the main criteria for evaluation.

The products produced may be accompanied by a marketing plan, so students can sell it later and earn some profit.

5. Students may interview professionals in various fields of study (e.g., economics, engineering, architecture, social sciences, etc. to find out how they make use of polynomials in their work.

As a final project, they may work with these professionals with the supervision of the teacher or some other adult or student’s guardian. In this way, students may have more appreciation for mathematics, and develop work ethics at the same time.

Assessment

In a scale of 6, where 6 is the highest score and 1 is the lowest score, rate the overall performance of the students in the culminating activity.

6. Exemplary response

➢ gives a complete response with a clear, coherent, unambiguous and elegant explanation;

➢ includes a clear and simplified diagram;

➢ communicates effectively to the identified audience;

➢ shows understanding of the open-ended problem's mathematical ideas and processes;

➢ identifies all the important elements of the problem;

➢ may include examples and counterexamples;

➢ presents strong supporting arguments.

5 Competent response

➢ gives a fairly complete response with reasonably clear explanations;

➢ may include an appropriate diagram;

➢ communicates effectively to the identified audience;

➢ shows understanding of the problem's mathematical ideas and

processes;

➢ identifies the most important elements of the problem;

➢ presents solid supporting arguments.

4 Minor Flaws But Satisfactory

➢ completes the problem satisfactorily, but the explanation may be muddled;

➢ argumentation may be incomplete; diagram may be inappropriate or unclear;

➢ understands the underlying mathematical ideas;

➢ uses mathematical ideas effectively.

3 Serious Flaws But Nearly Satisfactory

➢ begins the problem appropriately but may fail to complete or may omit significant parts of the problem;

➢ may fail to show full understanding of mathematical ideas and processes;

➢ may make major computational errors;

➢ may misuse or fail to use mathematical terms;

➢ response may reflect an inappropriate strategy for solving the problem.

2 Begins, But Fails to Complete Problem

➢ explanation is not understandable;

➢ diagram may be unclear;

➢ shows no understanding of the problem situation;

➢ may make major computational errors.

1 Unable to Begin Effectively

➢ words do not reflect the problem;

➢ drawings misrepresent the problem situation;

➢ copies parts of the problem but without attempting a solution;

➢ fails to indicate which information is appropriate to then problem.

Source

-----------------------

Addition

are performed and expressed using

Subtraction

Multiplication

Evaluation of polynomials

by substitution

are explained

through

Division

Operations on polynomials

can be simplified through

in the form of

POLYNOMIALS

Algebraic Expressions

Terms and degrees of polynomials

Algorithm

leads to

Remainder

Theorem

Graphing of polynomial functions

can be illustrated by

Factor Theorem

Polynomial functions

P(x) = AnXn+An-1Xn-1+…+ A1X+A0

Zeroes of polynomial functions

Rational zeroes

Polynomial inequalities

Set B

This machine multiplies every input(x) by 3

x = 2

6

input

The value of the polynomial function is 6.

1

2

3

4

5

3

6

1

2

7

4

8

9

10

1

1

1

4

4

1

8

0

3

7

7

1

2

1

8

4

3

9

1

6

1

4

9

4

[pic]

[pic]

[pic]

r. 5

r8

6x – 4

Remainder is 8

Divisor is x - 2

Note: Emphasize that they should always place zero for all the missing terms of the polynomial.

[pic]

(divide the expression by 3 so we can use synthetic division)

3a2 + 5a + 6

p(x) = x3 + x2 – x - 1

[pic]

[pic]

[pic]

Sample Graphs of Linear Functions

y=x2

f(x) = x4 – x3 – 7x2 + x + 6

[pic] = x3 – 7x - 6

[pic] = x2 – x – 6

f(x) = x4 + 2x3 – 3x2 – 8x - 4

[pic]

(we use the remaining possible numbersto check if they are roots)

[pic]

-1 1

-1

[pic]

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