Teacher Road Map for Lesson 21: The Remainder Theorem · · 2017-07-16Teacher Road Map for Lesson 21: The Remainder Theorem . ... Examples: 𝑔𝑔(𝑥𝑥 ... In order for x

Hart Interactive – Algebra 2 M1 Lesson 21 ALGEBRA II

T

Teacher Road Map for Lesson 21: The Remainder Theorem

Objective in Student Friendly Language:

Today I am: comparing the graph of a polynomial to division of a factor

So that I can: understand how the remainder can be used to find specific points on a polynomial graph.

I’ll know I have it when I can: write a polynomial function bases on the roots and y-intercept.

Flow: o Students examine a polynomial function’s graph and equation to determine roots. They then

divide the polynomial by various binomials to see the relationship between the remainder and the point on the graph.

o Students review what they learned in Unit 1 to graph a 4th degree polynomial function given one root and refine their graph using the ideas of the Remainder Theorem.

o Students state the Remainder Theorem and then practice using it. o Students write an equation of a polynomial function based on the graph, roots and y-

intercept. Big Ideas:

o The Remainder Theorem is extremely useful for finding specific values of a polynomial function and for further defining graphs of polynomials. In later math courses, the Remainder Theorem helps limit the domain in calculus problems.

CCS Standards:

o A-APR.2 Know and apply the Remainder Theorem: For a polynomial f(x) and a number a, the remainder on division by x – a is p(a), so p(a) = 0 if and only if (x – a) is a factor of p(x).

Lesson 21: The Remainder Theorem

293

This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015

This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.

http://creativecommons.org/licenses/by-nc-sa/3.0/deed.en_US





T

Student Outcomes Students know and apply the remainder theorem and understand the role zeros play in the theorem.

Lesson Notes

In this lesson, students are primarily working on exercises that lead them to the concept of the remainder theorem, the connection between factors and zeros of a polynomial, and how this relates to the graph of a polynomial function. Students should understand that for a polynomial function 𝑃𝑃 and a number 𝑎𝑎, the remainder on division by 𝑥𝑥 − 𝑎𝑎 is the value 𝑃𝑃(𝑎𝑎) and extend this to the idea that 𝑃𝑃(𝑎𝑎) = 0 if and only if (𝑥𝑥 −𝑎𝑎) is a factor of the polynomial (A-APR.2). There should be plenty of discussion after each exercise.

Exploratory Exercise 1. Consider the polynomial 𝑷𝑷(𝒙𝒙) = 𝒙𝒙𝟒𝟒 + 𝟑𝟑𝒙𝒙𝟑𝟑 − 𝟐𝟐𝟐𝟐𝒙𝒙𝟐𝟐 − 𝟑𝟑𝟑𝟑𝒙𝒙 + 𝟏𝟏𝟒𝟒𝟒𝟒

shown graphed on the right. A. Is 𝟏𝟏 a zero of the polynomial 𝑷𝑷? How do you know?

No, f(1) ≠ 0. OR No, the function does not cross or touch the x-axis at x = 1.

B. Divide P(x) by x – 1. You may use any method (long division,

reserve tabular method or synthetic division). What is the remainder?

The remainder is 84. C. What is f(1)? How does this answer compare to the remainder

in Part B?

f(1) = 84. They are the same. D. Is 𝒙𝒙 + 𝟑𝟑 one of the factors of 𝑷𝑷? How do you know?

Yes x + 3 is a factor of P since the function crosses the x-axis at -3. E. Divide P(x) by x + 3. What is the remainder?

P(x) ÷ (x + 3) = x3 – 28x + 48. The remainder is 0. F. What is f(-3)? How does this answer compare to the remainder in Part E?

f(-3) = 0. They are the same.


294








T

2. The polynomial function, 4 3 2( ) 7 6f x x x x x= + − − + , has a zero at -3. Graph the function on the grid at the right.

A. Use what you learned in Exercise 1 and your knowledge of

polynomials from Unit 1 to create an accurate graph of the function. B. What is the value of f(-2)? Mark this point on your graph.

f(-2) = -12. C. What is the value of f(0)? Mark this point on your graph.

f(0) = 6. D. What is the value of f(-4)? Can this be marked on your graph?

f(-4) = 90. It cannot be marked on the graph since we don’t have values that great.

Discussion 3. The Remainder Theorem states: If the polynomial P(x) is divided by x – a, then the remainder is

__P(a)_ ___. 4. Why is this important? How this can help us graph polynomial functions?

Answers will vary. By allowing us a different way to find coordinate points on a polynomial graph we can sketch the graph more accurately.

-14

-13

-12

-11

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

-4 -3 -2 -1 0 1 2 3 4


295








T

Exercises 5–7 (5 minutes)

Assign different groups of students one of the three problems. Have them complete their assigned problem, and then have a student from each group put their solution on the board. Having the solutions readily available allows students to start looking for a pattern without making the lesson too tedious.

5. Consider the polynomial function 𝒇𝒇(𝒙𝒙) = 𝟑𝟑𝒙𝒙𝟐𝟐 + 𝟐𝟐𝒙𝒙 − 𝟒𝟒.

A. Divide 𝒇𝒇 by 𝒙𝒙 − 𝟐𝟐.

B. Find 𝒇𝒇(𝟐𝟐).

f(x)x − 2

= 3x2 + 8x − 4

x − 2

= (3x + 14) +24

x − 2

f(2) = 24

6. Consider the polynomial function 𝒈𝒈(𝒙𝒙) = 𝒙𝒙𝟑𝟑 − 𝟑𝟑𝒙𝒙𝟐𝟐 + 𝟑𝟑𝒙𝒙 + 𝟐𝟐.

A. Divide 𝒈𝒈 by 𝒙𝒙 + 𝟏𝟏. B. Find 𝒈𝒈(−𝟏𝟏).

g(x)x + 1

=x3 − 3x2 + 6x + 8

x + 1

= (x2 − 4x + 10)−2

x + 1

g(−1) = −2

7. Consider the polynomial function 𝒉𝒉(𝒙𝒙) = 𝒙𝒙𝟑𝟑 + 𝟐𝟐𝒙𝒙 − 𝟑𝟑.

A. Divide 𝒉𝒉 by 𝒙𝒙 − 𝟑𝟑. B. Find 𝒉𝒉(𝟑𝟑).

h(x)x − 3

=x3 + 2x − 3

x − 3

= (x2 + 3x + 11) +30

x − 3

h(3) = 30

Scaffolding: If students are struggling, replace the polynomials in Exercises 6 and 7 with easier polynomial functions.

Examples:

𝑔𝑔(𝑥𝑥) = 𝑥𝑥2 − 7𝑥𝑥 − 11 a. Divide by 𝑥𝑥 + 1. b. Find 𝑔𝑔(−1).

(𝑥𝑥 − 8) −3

𝑥𝑥 + 1 𝑔𝑔(−1) = −3

ℎ(𝑥𝑥) = 2𝑥𝑥2 + 9 a. Divide by 𝑥𝑥 − 3. b. Find ℎ(3).

(2𝑥𝑥 + 6) +27

2𝑥𝑥2 + 9 ℎ(3) = 27


296








T

Discussion (7 minutes)

What is 𝑓𝑓(2)? What is 𝑔𝑔(−1)? What is ℎ(3)? 𝑓𝑓(2) = 24; 𝑔𝑔(−1) = −2; ℎ(3) = 30

Looking at the results of the quotient, what pattern do we see? The remainder is the value of the function.

Stating this in more general terms, what do we suspect about the connection between the remainder from dividing a polynomial 𝑃𝑃 by 𝑥𝑥 − 𝑎𝑎 and the value of 𝑃𝑃(𝑎𝑎)? The remainder found after dividing 𝑃𝑃 by 𝑥𝑥 − 𝑎𝑎 will be the same value as 𝑃𝑃(𝑎𝑎).

Why would this be? Think about the quotient 133

. We could write this as 13 = 4 ∙ 3 + 1, where 4 is the quotient and 1 is the remainder.

Apply this same principle to Exercise 1. Write the following on the board, and talk through it:

𝑓𝑓(𝑥𝑥)𝑥𝑥 − 2

=3𝑥𝑥2 + 8𝑥𝑥 − 4

𝑥𝑥 − 2= (3𝑥𝑥 + 14) +

24𝑥𝑥 − 2

How can we rewrite 𝑓𝑓 using the equation above? Multiply both sides of the equation by 𝑥𝑥 − 2 to get 𝑓𝑓(𝑥𝑥) = (3𝑥𝑥 + 14)(𝑥𝑥 − 2) + 24.

In general we can say that if you divide polynomial 𝑃𝑃 by 𝑥𝑥 − 𝑎𝑎, then the remainder must be a number; in fact, there is a (possibly non-zero degree) polynomial function 𝑞𝑞 such that the equation,

𝑃𝑃(𝑥𝑥) = 𝒒𝒒(𝒙𝒙) ∙ (𝑥𝑥 − 𝑎𝑎) + 𝒓𝒓

↑ ↑ quotient remainder

is true for all 𝑥𝑥. What is 𝑃𝑃(𝑎𝑎)?

𝑃𝑃(𝑎𝑎) = 𝑞𝑞(𝑎𝑎)(𝑎𝑎 − 𝑎𝑎) + 𝑟𝑟 = 𝑞𝑞(𝑎𝑎) ∙ 0 + 𝑟𝑟 = 0 + 𝑟𝑟 = 𝑟𝑟 We have just proven the remainder theorem, which is formally stated in the box below.

Restate the remainder theorem in your own words to your partner.

While students are doing this, circulate and informally assess student understanding before asking students to share their responses as a class.

REMAINDER THEOREM: Let 𝑃𝑃 be a polynomial function in 𝑥𝑥, and let 𝑎𝑎 be any real number. Then there exists a unique polynomial function 𝑞𝑞 such that the equation

𝑃𝑃(𝑥𝑥) = 𝑞𝑞(𝑥𝑥)(𝑥𝑥 − 𝑎𝑎) + 𝑃𝑃(𝑎𝑎) is true for all 𝑥𝑥. That is, when a polynomial is divided by (𝑥𝑥 − 𝑎𝑎), the remainder is the value of the polynomial evaluated at 𝑎𝑎.

MP.8


297








T

Exercise 8 (5 minutes)

Students may need more guidance through this exercise, but allow them to struggle with it first. After a few students have found 𝑘𝑘, share various methods used.

8. Consider the polynomial 𝑷𝑷(𝒙𝒙) = 𝒙𝒙𝟑𝟑 + 𝒌𝒌𝒙𝒙𝟐𝟐 + 𝒙𝒙 + 𝟑𝟑. A. Find the value of 𝒌𝒌 so that 𝒙𝒙 + 𝟏𝟏 is a factor of 𝑷𝑷.

In order for x + 1 to be a factor of P, the remainder must be zero. Hence, since x + 1 = x − (−1), we must have P(−1) = 0 so that 0 = −1 + k − 1 +6. Then k = −4.

B. Find the other two factors of 𝑷𝑷 for the value of 𝒌𝒌 found in Part A

𝑃𝑃(𝑥𝑥) = (𝑥𝑥 + 1)(𝑥𝑥2 − 5𝑥𝑥 + 6) = (𝑥𝑥 + 1)(𝑥𝑥 − 2)(𝑥𝑥 − 3)

Discussion (7 minutes)

Remember that for any polynomial function 𝑃𝑃 and real number 𝑎𝑎, the remainder theorem says that there exists a polynomial 𝑞𝑞 so that 𝑃𝑃(𝑥𝑥) = 𝑞𝑞(𝑥𝑥)(𝑥𝑥 − 𝑎𝑎) + 𝑃𝑃(𝑎𝑎).

What does it mean if 𝑎𝑎 is a zero of a polynomial 𝑃𝑃? 𝑃𝑃(𝑎𝑎) = 0

So what does the remainder theorem say if 𝑎𝑎 is a zero of 𝑃𝑃? There is a polynomial 𝑞𝑞 so that 𝑃𝑃(𝑥𝑥) = 𝑞𝑞(𝑥𝑥)(𝑥𝑥 − 𝑎𝑎) + 0.

How does (𝑥𝑥 − 𝑎𝑎) relate to 𝑃𝑃 if 𝑎𝑎 is a zero of 𝑃𝑃? If 𝑎𝑎 is a zero of 𝑃𝑃, then (𝑥𝑥 − 𝑎𝑎) is a factor of 𝑃𝑃.

How does the graph of a polynomial function 𝑦𝑦 = 𝑃𝑃(𝑥𝑥) correspond to the equation of the polynomial 𝑃𝑃? The zeros are the 𝑥𝑥-intercepts of the graph of 𝑃𝑃. If we know a zero of 𝑃𝑃, then we know a

factor of 𝑃𝑃. If we know all of the zeros of a polynomial function, and their multiplicities, do we know the

equation of the function? Not necessarily. It is possible that the equation of the function contains some factors that

cannot factor into linear terms.

Scaffolding: Challenge early finishers with this problem:

Given that 𝑥𝑥 + 1 and 𝑥𝑥 − 1 are factors of 𝑃𝑃(𝑥𝑥) = 𝑥𝑥4 + 2𝑥𝑥3 −49𝑥𝑥2 − 2𝑥𝑥 + 48, write 𝑃𝑃 in factored form.

Answer: (𝑥𝑥 + 1)(𝑥𝑥 − 1)(𝑥𝑥 + 8)(𝑥𝑥 − 6)


298








T

We have just proven the factor theorem, which is a direct consequence of the remainder theorem.

Give an example of a polynomial function with zeros of multiplicity 2 at 1 and 3.

𝑃𝑃(𝑥𝑥) = (𝑥𝑥 − 1)2(𝑥𝑥 − 3)2 Give another example of a polynomial function with zeros of multiplicity 2 at 1 and 3.

𝑄𝑄(𝑥𝑥) = (𝑥𝑥 − 1)2(𝑥𝑥 − 3)2(𝑥𝑥2 + 1) or 𝑅𝑅(𝑥𝑥) = 4(𝑥𝑥 − 1)2(𝑥𝑥 − 3)2 If we know the zeros of a polynomial, does the factor theorem tell us the exact formula for the

polynomial? No. But, if we know the degree of the polynomial and the leading coefficient, we can often

deduce the equation of the polynomial.

Exercise 9 (6 minutes)

Allow students a few minutes to work on the problem and then share results.

9. Consider the graph of a degree 𝟓𝟓 polynomial shown to the right, with 𝒙𝒙-intercepts −𝟒𝟒, −𝟐𝟐, 𝟏𝟏, 𝟑𝟑, and 𝟓𝟓.

A. Write a formula for a possible polynomial function that the graph represents using 𝒄𝒄 as the constant factor.

𝑷𝑷(𝒙𝒙) = 𝒄𝒄(𝒙𝒙 + 𝟒𝟒)(𝒙𝒙 + 𝟐𝟐)(𝒙𝒙 − 𝟏𝟏)(𝒙𝒙 − 𝟑𝟑)(𝒙𝒙 − 𝟓𝟓)

B. Suppose the 𝒚𝒚-intercept is −𝟒𝟒. Find the value of 𝒄𝒄 so that the

graph of 𝑷𝑷 has 𝒚𝒚-intercept −𝟒𝟒.

𝑷𝑷(𝒙𝒙) =𝟏𝟏𝟑𝟑𝟑𝟑

(𝒙𝒙 + 𝟒𝟒)(𝒙𝒙 + 𝟐𝟐)(𝒙𝒙 − 𝟏𝟏)(𝒙𝒙 − 𝟑𝟑)(𝒙𝒙 − 𝟓𝟓)

Discussion Points What information from the graph was needed to write the equation?

The 𝑥𝑥-intercepts were needed to write the factors. Why would there be more than one polynomial function possible?

Because the factors could be multiplied by any constant and still produce a graph with the same 𝑥𝑥-intercepts.

FACTOR THEOREM: Let 𝑃𝑃 be a polynomial function in 𝑥𝑥, and let 𝑎𝑎 be any real number. If 𝑎𝑎 is a zero of 𝑃𝑃 then (𝑥𝑥 − 𝑎𝑎) is a factor of 𝑃𝑃.


299








T

Why can’t we find the constant factor 𝑐𝑐 by just knowing the zeros of the polynomial? The zeros only determine where the graph crosses the 𝑥𝑥-axis, not how the graph is stretched

vertically. The constant factor can be used to vertically scale the graph of the polynomial function that we found to fit the depicted graph.

Lesson Summary

REMAINDER THEOREM: Let 𝑃𝑃 be a polynomial function in 𝑥𝑥, and let 𝑎𝑎 be any real number. Then there exists a unique polynomial function 𝑞𝑞 such that the equation

𝑃𝑃(𝑥𝑥) = 𝑞𝑞(𝑥𝑥)(𝑥𝑥 − 𝑎𝑎) + 𝑃𝑃(𝑎𝑎)

is true for all 𝑥𝑥. That is, when a polynomial is divided by (𝑥𝑥 − 𝑎𝑎), the remainder is the value of the polynomial evaluated at 𝑎𝑎.

FACTOR THEOREM: Let 𝑃𝑃 be a polynomial function in 𝑥𝑥, and let 𝑎𝑎 be any real number. If 𝑎𝑎 is a zero of 𝑃𝑃, then (𝑥𝑥 − 𝑎𝑎) is a factor of 𝑃𝑃.

Example: If 𝑃𝑃(𝑥𝑥) = 𝑥𝑥3 − 5𝑥𝑥2 + 3𝑥𝑥 + 9, then 𝑃𝑃(3) = 27 − 45 + 9 + 9 = 0, so (𝑥𝑥 − 3) is a factor of 𝑃𝑃.

Closing (2 minutes)

Have students summarize the results of the remainder theorem and the factor theorem. What is the connection between the remainder when a polynomial 𝑃𝑃 is divided by 𝑥𝑥 − 𝑎𝑎 and the

value of 𝑃𝑃(𝑎𝑎)? They are the same.

If 𝑥𝑥 − 𝑎𝑎 is factor, then _________. The number 𝑎𝑎 is a zero of 𝑃𝑃.

If 𝑃𝑃(𝑎𝑎) = 0, then ____________. (𝑥𝑥 − 𝑎𝑎) is a factor of 𝑃𝑃.

Exit Ticket (5 minutes)


300








T

Name Date


Exit Ticket

Consider the polynomial 𝑃𝑃(𝑥𝑥) = 𝑥𝑥3 + 𝑥𝑥2 − 10𝑥𝑥 − 10.

1. Is 𝑥𝑥 + 1 one of the factors of 𝑃𝑃? Explain.

2. The graph shown has 𝑥𝑥-intercepts at √10, −1, and −√10. Could this be the graph of 𝑃𝑃(𝑥𝑥) = 𝑥𝑥3 + 𝑥𝑥2 − 10𝑥𝑥 − 10? Explain how you know.


301








T

Exit Ticket Sample Solutions

Consider polynomial 𝑷𝑷(𝒙𝒙) = 𝒙𝒙𝟑𝟑 + 𝒙𝒙𝟐𝟐 − 𝟏𝟏𝟑𝟑𝒙𝒙 − 𝟏𝟏𝟑𝟑.

1. Is 𝒙𝒙 + 𝟏𝟏 one of the factors of 𝑷𝑷? Explain.

𝑷𝑷(−𝟏𝟏) = (−𝟏𝟏)𝟑𝟑 + (−𝟏𝟏)𝟐𝟐 − 𝟏𝟏𝟑𝟑(−𝟏𝟏) − 𝟏𝟏𝟑𝟑 = −𝟏𝟏+ 𝟏𝟏 + 𝟏𝟏𝟑𝟑 − 𝟏𝟏𝟑𝟑 = 𝟑𝟑

Yes, 𝒙𝒙 + 𝟏𝟏 is a factor of 𝑷𝑷 because 𝑷𝑷(−𝟏𝟏) = 𝟑𝟑. Or, using factoring by grouping, we have

𝑷𝑷(𝒙𝒙) = 𝒙𝒙𝟐𝟐(𝒙𝒙 + 𝟏𝟏) − 𝟏𝟏𝟑𝟑(𝒙𝒙 + 𝟏𝟏) = (𝒙𝒙 + 𝟏𝟏)(𝒙𝒙𝟐𝟐 − 𝟏𝟏𝟑𝟑).

2. The graph shown has 𝒙𝒙-intercepts at √𝟏𝟏𝟑𝟑, −𝟏𝟏, and −√𝟏𝟏𝟑𝟑. Could this be the graph of 𝑷𝑷(𝒙𝒙) = 𝒙𝒙𝟑𝟑 + 𝒙𝒙𝟐𝟐 −𝟏𝟏𝟑𝟑𝒙𝒙 − 𝟏𝟏𝟑𝟑? Explain how you know.

Yes, this could be the graph of 𝑷𝑷. Since this graph has 𝒙𝒙-intercepts at √𝟏𝟏𝟑𝟑, −𝟏𝟏, and −√𝟏𝟏𝟑𝟑, the factor theorem says that (𝒙𝒙 − √𝟏𝟏𝟑𝟑), (𝒙𝒙 − 𝟏𝟏), and (𝒙𝒙 +√𝟏𝟏𝟑𝟑 ) are all factors of the equation that goes with this graph. Since �𝒙𝒙 − √𝟏𝟏𝟑𝟑��𝒙𝒙 + √𝟏𝟏𝟑𝟑�(𝒙𝒙 − 𝟏𝟏) = 𝒙𝒙𝟑𝟑 + 𝒙𝒙𝟐𝟐 − 𝟏𝟏𝟑𝟑𝒙𝒙 − 𝟏𝟏𝟑𝟑, the graph shown is quite likely to be the graph of 𝑷𝑷.

Homework Problem Set Sample Solutions

1. Use the remainder theorem to find the remainder for each of the following divisions.

a. �𝒙𝒙𝟐𝟐+𝟑𝟑𝒙𝒙+𝟏𝟏�

(𝒙𝒙+𝟐𝟐) b. 𝒙𝒙𝟑𝟑−𝟑𝟑𝒙𝒙𝟐𝟐−𝟕𝟕𝒙𝒙+𝟗𝟗

(𝒙𝒙−𝟑𝟑)

−𝟏𝟏 −𝟑𝟑𝟗𝟗

c. 𝟑𝟑𝟐𝟐𝒙𝒙𝟒𝟒+𝟐𝟐𝟒𝟒𝒙𝒙𝟑𝟑−𝟏𝟏𝟐𝟐𝒙𝒙𝟐𝟐+𝟐𝟐𝒙𝒙+𝟏𝟏

(𝒙𝒙+𝟏𝟏)

−𝟓𝟓

d. 𝟑𝟑𝟐𝟐𝒙𝒙𝟒𝟒+𝟐𝟐𝟒𝟒𝒙𝒙𝟑𝟑−𝟏𝟏𝟐𝟐𝒙𝒙𝟐𝟐+𝟐𝟐𝒙𝒙+𝟏𝟏

(𝟐𝟐𝒙𝒙−𝟏𝟏)

Hint for part (d): Can you rewrite the division expression so that the divisor is in the form (𝒙𝒙 − 𝒄𝒄) for some constant 𝒄𝒄?

𝟒𝟒


302








T

2. Consider the polynomial 𝑷𝑷(𝒙𝒙) = 𝒙𝒙𝟑𝟑 + 𝟑𝟑𝒙𝒙𝟐𝟐 − 𝟐𝟐𝒙𝒙 − 𝟏𝟏. Find 𝑷𝑷(𝟒𝟒) in two ways.

𝑃𝑃(4) = 43 + 6(4)2 − 8(4) − 1 = 127

𝑥𝑥3+6𝑥𝑥2−8𝑥𝑥−1

𝑥𝑥−4 has a remainder of 127, so 𝑃𝑃(4) = 127.

3. Consider the polynomial function 𝑷𝑷(𝒙𝒙) = 𝟐𝟐𝒙𝒙𝟒𝟒 + 𝟑𝟑𝒙𝒙𝟐𝟐 + 𝟏𝟏𝟐𝟐. a. Divide 𝑷𝑷 by 𝒙𝒙 + 𝟐𝟐, and rewrite 𝑷𝑷 in the form (divisor)(quotient)+remainder.

𝑃𝑃(𝑥𝑥) = (𝑥𝑥 + 2)(2𝑥𝑥3 − 4𝑥𝑥2 + 11𝑥𝑥 − 22) + 56 b. Find 𝑷𝑷(−𝟐𝟐).

𝑃𝑃(−2) = (−2 + 2)�𝑞𝑞(−2)� + 56 = 56

4. Consider the polynomial function 𝑷𝑷(𝒙𝒙) = 𝒙𝒙𝟑𝟑 + 𝟒𝟒𝟐𝟐. a. Divide 𝑷𝑷 by 𝒙𝒙 − 𝟒𝟒, and rewrite 𝑷𝑷 in the form (divisor)(quotient) + remainder.

𝑃𝑃(𝑥𝑥) = (𝑥𝑥 − 4)(𝑥𝑥2 + 4𝑥𝑥 + 16) + 106

b. Find 𝑷𝑷(𝟒𝟒).

𝑃𝑃(4) = (4 − 4)�𝑞𝑞(4)� + 106 = 106

5. Explain why for a polynomial function 𝑷𝑷, 𝑷𝑷(𝒂𝒂) is equal to the remainder of the quotient of 𝑷𝑷

and 𝒙𝒙 − 𝒂𝒂.

The polynomial P can be rewritten in the form P(x) = (x − a)(q(x)) + r, where q(x) is the quotient function and r is the remainder. Then P(a) = (a − a)�q(a)� + r. Therefore, P(a) = r.

6. Is 𝒙𝒙 − 𝟓𝟓 a factor of the function 𝒇𝒇(𝒙𝒙) = 𝒙𝒙𝟑𝟑 + 𝒙𝒙𝟐𝟐 − 𝟐𝟐𝟕𝟕𝒙𝒙 − 𝟏𝟏𝟓𝟓? Show work supporting your answer.

Yes, because f(5) = 0 means that dividing by x − 5 leaves a remainder of 0.


303








T

7. Is 𝒙𝒙 + 𝟏𝟏 a factor of the function 𝒇𝒇(𝒙𝒙) = 𝟐𝟐𝒙𝒙𝟓𝟓 − 𝟒𝟒𝒙𝒙𝟒𝟒 + 𝟗𝟗𝒙𝒙𝟑𝟑 − 𝒙𝒙 + 𝟏𝟏𝟑𝟑? Show work supporting your answer.

No, because f(−1) = −1 means that dividing by x + 1 has a remainder of −1.

8. A polynomial function 𝒑𝒑 has zeros of 𝟐𝟐, 𝟐𝟐, −𝟑𝟑, −𝟑𝟑, −𝟑𝟑, and 𝟒𝟒. Find a possible formula for 𝑷𝑷,

and state its degree. Why is the degree of the polynomial not 𝟑𝟑?

One solution is P(x) = (x − 2)2(x + 3)3(x − 4). The degree of P is 6. This is not a degree 3 polynomial function because the factor (x − 2) appears twice, and the factor (x + 3) appears 3 times, while the factor (x − 4) appears once.

9. Consider the polynomial function 𝑷𝑷(𝒙𝒙) = 𝒙𝒙𝟑𝟑 + 𝟐𝟐𝒙𝒙𝟐𝟐 − 𝟏𝟏𝟓𝟓𝒙𝒙. a. Verify that 𝑷𝑷(𝟑𝟑) = 𝟑𝟑. Since 𝑷𝑷(𝟑𝟑) = 𝟑𝟑, what must one of

the factors of 𝑷𝑷 be?

𝑃𝑃(0) = 903 + 2(02)− 15(0)0 = 0; 𝑥𝑥 − 0 or x.

b. Find the remaining two factors of 𝑷𝑷.

𝑷𝑷(𝒙𝒙) = 𝒙𝒙(𝒙𝒙 − 𝟑𝟑)(𝒙𝒙 + 𝟓𝟓)

c. State the zeros of 𝑷𝑷.

𝒙𝒙 = 𝟑𝟑, 𝟑𝟑, −𝟓𝟓

d. Sketch the graph of 𝑷𝑷.


304








T

10. Consider the polynomial function 𝑷𝑷(𝒙𝒙) = 𝟐𝟐𝒙𝒙𝟑𝟑 + 𝟑𝟑𝒙𝒙𝟐𝟐 − 𝟐𝟐𝒙𝒙 − 𝟑𝟑. a. Verify that 𝑷𝑷(−𝟏𝟏) = 𝟑𝟑. Since 𝑷𝑷(−𝟏𝟏) = 𝟑𝟑, what must one of the factors of 𝑷𝑷 be?

𝑃𝑃(−1) = 2(−1)3 + 3(−1)2 − 2(−1)− 3 = 0; 𝑥𝑥 + 1

b. Find the remaining two factors of 𝑷𝑷.

𝑃𝑃(𝑥𝑥) = (𝑥𝑥 + 1)(𝑥𝑥 − 1)(2𝑥𝑥 + 3)

c. State the zeros of 𝑷𝑷.

𝑥𝑥 = −1, 1, −32

d. Sketch the graph of 𝑷𝑷.

11. The graph to the right is of a third-degree polynomial

function 𝒇𝒇. a. State the zeros of 𝒇𝒇.

𝑥𝑥 = −10, − 1, 2

b. Write a formula for 𝒇𝒇 in factored form using 𝒄𝒄 for the

constant factor.

𝑓𝑓(𝑥𝑥) = 𝑐𝑐(𝑥𝑥 + 10)(𝑥𝑥 + 1)(𝑥𝑥 − 2)

c. Use the fact that 𝒇𝒇(−𝟒𝟒) = −𝟓𝟓𝟒𝟒 to find the constant factor 𝒄𝒄.

−54 = 𝑐𝑐(−4 + 10)(−4 + 1)(−4− 2)

𝑐𝑐 = −12

𝑓𝑓(𝑥𝑥) = −12

(𝑥𝑥 + 10)(𝑥𝑥 + 1)(𝑥𝑥 − 2)

d. Verify your equation by using the fact that 𝒇𝒇(𝟏𝟏) = 𝟏𝟏𝟏𝟏.

𝑓𝑓(1) = −12

(1 + 10)(1 + 1)(1 − 2) = −12

(11)(2)(−1) = 11


305








T

12. Find the value of 𝒌𝒌 so that 𝒙𝒙𝟑𝟑−𝒌𝒌𝒙𝒙𝟐𝟐+𝟐𝟐𝒙𝒙−𝟏𝟏

has remainder 𝟐𝟐.

𝑘𝑘 = −5

13. Find the value 𝒌𝒌 so that 𝒌𝒌𝒙𝒙𝟑𝟑+𝒙𝒙−𝒌𝒌𝒙𝒙+𝟐𝟐

has remainder 𝟏𝟏𝟑𝟑.

𝑘𝑘 = −2

14. Show that 𝒙𝒙𝟓𝟓𝟏𝟏 − 𝟐𝟐𝟏𝟏𝒙𝒙 + 𝟐𝟐𝟑𝟑 is divisible by 𝒙𝒙 − 𝟏𝟏.

Let 𝑃𝑃(𝑥𝑥) = 𝑥𝑥51 − 21𝑥𝑥 + 20.

Then 𝑃𝑃(1) = 151 − 21(1) + 20 = 1 − 21 + 20 = 0.

Since 𝑃𝑃(1) = 0, the remainder of the quotient (𝑥𝑥51 − 21𝑥𝑥 + 20) ÷ (𝑥𝑥 − 1) is 0.

Therefore, 𝑥𝑥51 − 21𝑥𝑥 + 20 is divisible by 𝑥𝑥 − 1.

15. Show that 𝒙𝒙 + 𝟏𝟏 is a factor of 𝟏𝟏𝟗𝟗𝒙𝒙𝟒𝟒𝟐𝟐 + 𝟏𝟏𝟐𝟐𝒙𝒙 − 𝟏𝟏.

Let 𝑃𝑃(𝑥𝑥) = 19𝑥𝑥42 + 18𝑥𝑥 − 1.

Then 𝑃𝑃(−1) = 19(−1)42 + 18(−1)− 1 = 19 − 18 − 1 = 0.

Since 𝑃𝑃(−1) = 0, 𝑥𝑥 + 1 must be a factor of 𝑃𝑃.


306







Documents

Teacher Road Map for Lesson 21: The Remainder Theorem · · 2017-07-16Teacher Road Map for Lesson 21: The Remainder Theorem . ... Examples: 𝑔𝑔(𝑥𝑥 ... In order for x