Teacher Road Map for Lesson 21: The Remainder TheoremΒ Β· Β Β· 2017-07-16Teacher Road Map for Lesson...
Transcript of Teacher Road Map for Lesson 21: The Remainder TheoremΒ Β· Β Β· 2017-07-16Teacher Road Map for Lesson...
Hart Interactive β Algebra 2 M1 Lesson 21 ALGEBRA II
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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.
Iβll know I have it when I can: write a polynomial function bases on the roots and y-intercept.
Flow: o Students examine a polynomial functionβs 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 β a is p(a), so p(a) = 0 if and only if (x β a) is a factor of p(x).
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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 ππ and a number ππ, the remainder on division by π₯π₯ β ππ is the value ππ(ππ) and extend this to the idea that ππ(ππ) = 0 if and only if (π₯π₯ βππ) is a factor of the polynomial (A-APR.2). There should be plenty of discussion after each exercise.
Exploratory Exercise 1. Consider the polynomial π·π·(ππ) = ππππ + ππππππ β ππππππππ β ππππππ + ππππππ
shown graphed on the right. A. Is ππ a zero of the polynomial π·π·? How do you know?
No, f(1) β 0. OR No, the function does not cross or touch the x-axis at x = 1.
B. Divide P(x) by x β 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 ππ + ππ one of the factors of π·π·? 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) Γ· (x + 3) = x3 β 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.
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2. The polynomial function, 4 3 2( ) 7 6f x x x x x= + β β + , 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 donβt have values that great.
Discussion 3. The Remainder Theorem states: If the polynomial P(x) is divided by x β 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
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-1
0
1
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9
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13
14
-4 -3 -2 -1 0 1 2 3 4
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Exercises 5β7 (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 ππ(ππ) = ππππππ + ππππ β ππ.
A. Divide ππ by ππ β ππ.
B. Find ππ(ππ).
f(x)x β 2
= 3x2 + 8x β 4
x β 2
= (3x + 14) +24
x β 2
f(2) = 24
6. Consider the polynomial function ππ(ππ) = ππππ β ππππππ + ππππ + ππ.
A. Divide ππ by ππ + ππ. B. Find ππ(βππ).
g(x)x + 1
=x3 β 3x2 + 6x + 8
x + 1
= (x2 β 4x + 10)β2
x + 1
g(β1) = β2
7. Consider the polynomial function ππ(ππ) = ππππ + ππππ β ππ.
A. Divide ππ by ππ β ππ. B. Find ππ(ππ).
h(x)x β 3
=x3 + 2x β 3
x β 3
= (x2 + 3x + 11) +30
x β 3
h(3) = 30
Scaffolding: If students are struggling, replace the polynomials in Exercises 6 and 7 with easier polynomial functions.
Examples:
ππ(π₯π₯) = π₯π₯2 β 7π₯π₯ β 11 a. Divide by π₯π₯ + 1. b. Find ππ(β1).
(π₯π₯ β 8) β3
π₯π₯ + 1 ππ(β1) = β3
β(π₯π₯) = 2π₯π₯2 + 9 a. Divide by π₯π₯ β 3. b. Find β(3).
(2π₯π₯ + 6) +27
2π₯π₯2 + 9 β(3) = 27
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Discussion (7 minutes)
What is ππ(2)? What is ππ(β1)? What is β(3)? ππ(2) = 24; ππ(β1) = β2; β(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 ππ by π₯π₯ β ππ and the value of ππ(ππ)? The remainder found after dividing ππ by π₯π₯ β ππ will be the same value as ππ(ππ).
Why would this be? Think about the quotient 133
. We could write this as 13 = 4 β 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:
ππ(π₯π₯)π₯π₯ β 2
=3π₯π₯2 + 8π₯π₯ β 4
π₯π₯ β 2= (3π₯π₯ + 14) +
24π₯π₯ β 2
How can we rewrite ππ using the equation above? Multiply both sides of the equation by π₯π₯ β 2 to get ππ(π₯π₯) = (3π₯π₯ + 14)(π₯π₯ β 2) + 24.
In general we can say that if you divide polynomial ππ by π₯π₯ β ππ, then the remainder must be a number; in fact, there is a (possibly non-zero degree) polynomial function ππ such that the equation,
ππ(π₯π₯) = ππ(ππ) β (π₯π₯ β ππ) + ππ
β β quotient remainder
is true for all π₯π₯. What is ππ(ππ)?
ππ(ππ) = ππ(ππ)(ππ β ππ) + ππ = ππ(ππ) β 0 + ππ = 0 + ππ = ππ 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 ππ be a polynomial function in π₯π₯, and let ππ be any real number. Then there exists a unique polynomial function ππ such that the equation
ππ(π₯π₯) = ππ(π₯π₯)(π₯π₯ β ππ) + ππ(ππ) is true for all π₯π₯. That is, when a polynomial is divided by (π₯π₯ β ππ), the remainder is the value of the polynomial evaluated at ππ.
MP.8
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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 ππ, share various methods used.
8. Consider the polynomial π·π·(ππ) = ππππ + ππππππ + ππ + ππ. A. Find the value of ππ so that ππ + ππ is a factor of π·π·.
In order for x + 1 to be a factor of P, the remainder must be zero. Hence, since x + 1 = x β (β1), we must have P(β1) = 0 so that 0 = β1 + k β 1 +6. Then k = β4.
B. Find the other two factors of π·π· for the value of ππ found in Part A
ππ(π₯π₯) = (π₯π₯ + 1)(π₯π₯2 β 5π₯π₯ + 6) = (π₯π₯ + 1)(π₯π₯ β 2)(π₯π₯ β 3)
Discussion (7 minutes)
Remember that for any polynomial function ππ and real number ππ, the remainder theorem says that there exists a polynomial ππ so that ππ(π₯π₯) = ππ(π₯π₯)(π₯π₯ β ππ) + ππ(ππ).
What does it mean if ππ is a zero of a polynomial ππ? ππ(ππ) = 0
So what does the remainder theorem say if ππ is a zero of ππ? There is a polynomial ππ so that ππ(π₯π₯) = ππ(π₯π₯)(π₯π₯ β ππ) + 0.
How does (π₯π₯ β ππ) relate to ππ if ππ is a zero of ππ? If ππ is a zero of ππ, then (π₯π₯ β ππ) is a factor of ππ.
How does the graph of a polynomial function π¦π¦ = ππ(π₯π₯) correspond to the equation of the polynomial ππ? The zeros are the π₯π₯-intercepts of the graph of ππ. If we know a zero of ππ, then we know a
factor of ππ. 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 π₯π₯ + 1 and π₯π₯ β 1 are factors of ππ(π₯π₯) = π₯π₯4 + 2π₯π₯3 β49π₯π₯2 β 2π₯π₯ + 48, write ππ in factored form.
Answer: (π₯π₯ + 1)(π₯π₯ β 1)(π₯π₯ + 8)(π₯π₯ β 6)
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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.
ππ(π₯π₯) = (π₯π₯ β 1)2(π₯π₯ β 3)2 Give another example of a polynomial function with zeros of multiplicity 2 at 1 and 3.
ππ(π₯π₯) = (π₯π₯ β 1)2(π₯π₯ β 3)2(π₯π₯2 + 1) or π π (π₯π₯) = 4(π₯π₯ β 1)2(π₯π₯ β 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 ππ polynomial shown to the right, with ππ-intercepts βππ, βππ, ππ, ππ, and ππ.
A. Write a formula for a possible polynomial function that the graph represents using ππ as the constant factor.
π·π·(ππ) = ππ(ππ + ππ)(ππ + ππ)(ππ β ππ)(ππ β ππ)(ππ β ππ)
B. Suppose the ππ-intercept is βππ. Find the value of ππ so that the
graph of π·π· has ππ-intercept βππ.
π·π·(ππ) =ππππππ
(ππ + ππ)(ππ + ππ)(ππ β ππ)(ππ β ππ)(ππ β ππ)
Discussion Points What information from the graph was needed to write the equation?
The π₯π₯-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 π₯π₯-intercepts.
FACTOR THEOREM: Let ππ be a polynomial function in π₯π₯, and let ππ be any real number. If ππ is a zero of ππ then (π₯π₯ β ππ) is a factor of ππ.
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Why canβt we find the constant factor ππ by just knowing the zeros of the polynomial? The zeros only determine where the graph crosses the π₯π₯-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 ππ be a polynomial function in π₯π₯, and let ππ be any real number. Then there exists a unique polynomial function ππ such that the equation
ππ(π₯π₯) = ππ(π₯π₯)(π₯π₯ β ππ) + ππ(ππ)
is true for all π₯π₯. That is, when a polynomial is divided by (π₯π₯ β ππ), the remainder is the value of the polynomial evaluated at ππ.
FACTOR THEOREM: Let ππ be a polynomial function in π₯π₯, and let ππ be any real number. If ππ is a zero of ππ, then (π₯π₯ β ππ) is a factor of ππ.
Example: If ππ(π₯π₯) = π₯π₯3 β 5π₯π₯2 + 3π₯π₯ + 9, then ππ(3) = 27 β 45 + 9 + 9 = 0, so (π₯π₯ β 3) is a factor of ππ.
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 ππ is divided by π₯π₯ β ππ and the
value of ππ(ππ)? They are the same.
If π₯π₯ β ππ is factor, then _________. The number ππ is a zero of ππ.
If ππ(ππ) = 0, then ____________. (π₯π₯ β ππ) is a factor of ππ.
Exit Ticket (5 minutes)
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Name Date
Lesson 21: The Remainder Theorem
Exit Ticket
Consider the polynomial ππ(π₯π₯) = π₯π₯3 + π₯π₯2 β 10π₯π₯ β 10.
1. Is π₯π₯ + 1 one of the factors of ππ? Explain.
2. The graph shown has π₯π₯-intercepts at β10, β1, and ββ10. Could this be the graph of ππ(π₯π₯) = π₯π₯3 + π₯π₯2 β 10π₯π₯ β 10? Explain how you know.
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Exit Ticket Sample Solutions
Consider polynomial π·π·(ππ) = ππππ + ππππ β ππππππ β ππππ.
1. Is ππ + ππ one of the factors of π·π·? Explain.
π·π·(βππ) = (βππ)ππ + (βππ)ππ β ππππ(βππ) β ππππ = βππ+ ππ + ππππ β ππππ = ππ
Yes, ππ + ππ is a factor of π·π· because π·π·(βππ) = ππ. Or, using factoring by grouping, we have
π·π·(ππ) = ππππ(ππ + ππ) β ππππ(ππ + ππ) = (ππ + ππ)(ππππ β ππππ).
2. The graph shown has ππ-intercepts at βππππ, βππ, and ββππππ. Could this be the graph of π·π·(ππ) = ππππ + ππππ βππππππ β ππππ? Explain how you know.
Yes, this could be the graph of π·π·. Since this graph has ππ-intercepts at βππππ, βππ, and ββππππ, the factor theorem says that (ππ β βππππ), (ππ β ππ), and (ππ +βππππ ) are all factors of the equation that goes with this graph. Since οΏ½ππ β βπππποΏ½οΏ½ππ + βπππποΏ½(ππ β ππ) = ππππ + ππππ β ππππππ β ππππ, the graph shown is quite likely to be the graph of π·π·.
Homework Problem Set Sample Solutions
1. Use the remainder theorem to find the remainder for each of the following divisions.
a. οΏ½ππππ+ππππ+πποΏ½
(ππ+ππ) b. ππππβππππππβππππ+ππ
(ππβππ)
βππ βππππ
c. ππππππππ+ππππππππβππππππππ+ππππ+ππ
(ππ+ππ)
βππ
d. ππππππππ+ππππππππβππππππππ+ππππ+ππ
(ππππβππ)
Hint for part (d): Can you rewrite the division expression so that the divisor is in the form (ππ β ππ) for some constant ππ?
ππ
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2. Consider the polynomial π·π·(ππ) = ππππ + ππππππ β ππππ β ππ. Find π·π·(ππ) in two ways.
ππ(4) = 43 + 6(4)2 β 8(4) β 1 = 127
π₯π₯3+6π₯π₯2β8π₯π₯β1
π₯π₯β4 has a remainder of 127, so ππ(4) = 127.
3. Consider the polynomial function π·π·(ππ) = ππππππ + ππππππ + ππππ. a. Divide π·π· by ππ + ππ, and rewrite π·π· in the form (divisor)(quotient)+remainder.
ππ(π₯π₯) = (π₯π₯ + 2)(2π₯π₯3 β 4π₯π₯2 + 11π₯π₯ β 22) + 56 b. Find π·π·(βππ).
ππ(β2) = (β2 + 2)οΏ½ππ(β2)οΏ½ + 56 = 56
4. Consider the polynomial function π·π·(ππ) = ππππ + ππππ. a. Divide π·π· by ππ β ππ, and rewrite π·π· in the form (divisor)(quotient) + remainder.
ππ(π₯π₯) = (π₯π₯ β 4)(π₯π₯2 + 4π₯π₯ + 16) + 106
b. Find π·π·(ππ).
ππ(4) = (4 β 4)οΏ½ππ(4)οΏ½ + 106 = 106
5. Explain why for a polynomial function π·π·, π·π·(ππ) is equal to the remainder of the quotient of π·π·
and ππ β ππ.
The polynomial P can be rewritten in the form P(x) = (x β a)(q(x)) + r, where q(x) is the quotient function and r is the remainder. Then P(a) = (a β a)οΏ½q(a)οΏ½ + r. Therefore, P(a) = r.
6. Is ππ β ππ a factor of the function ππ(ππ) = ππππ + ππππ β ππππππ β ππππ? Show work supporting your answer.
Yes, because f(5) = 0 means that dividing by x β 5 leaves a remainder of 0.
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7. Is ππ + ππ a factor of the function ππ(ππ) = ππππππ β ππππππ + ππππππ β ππ + ππππ? Show work supporting your answer.
No, because f(β1) = β1 means that dividing by x + 1 has a remainder of β1.
8. A polynomial function ππ has zeros of ππ, ππ, βππ, βππ, βππ, and ππ. Find a possible formula for π·π·,
and state its degree. Why is the degree of the polynomial not ππ?
One solution is P(x) = (x β 2)2(x + 3)3(x β 4). The degree of P is 6. This is not a degree 3 polynomial function because the factor (x β 2) appears twice, and the factor (x + 3) appears 3 times, while the factor (x β 4) appears once.
9. Consider the polynomial function π·π·(ππ) = ππππ + ππππππ β ππππππ. a. Verify that π·π·(ππ) = ππ. Since π·π·(ππ) = ππ, what must one of
the factors of π·π· be?
ππ(0) = 903 + 2(02)β 15(0)0 = 0; π₯π₯ β 0 or x.
b. Find the remaining two factors of π·π·.
π·π·(ππ) = ππ(ππ β ππ)(ππ + ππ)
c. State the zeros of π·π·.
ππ = ππ, ππ, βππ
d. Sketch the graph of π·π·.
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10. Consider the polynomial function π·π·(ππ) = ππππππ + ππππππ β ππππ β ππ. a. Verify that π·π·(βππ) = ππ. Since π·π·(βππ) = ππ, what must one of the factors of π·π· be?
ππ(β1) = 2(β1)3 + 3(β1)2 β 2(β1)β 3 = 0; π₯π₯ + 1
b. Find the remaining two factors of π·π·.
ππ(π₯π₯) = (π₯π₯ + 1)(π₯π₯ β 1)(2π₯π₯ + 3)
c. State the zeros of π·π·.
π₯π₯ = β1, 1, β32
d. Sketch the graph of π·π·.
11. The graph to the right is of a third-degree polynomial
function ππ. a. State the zeros of ππ.
π₯π₯ = β10, β 1, 2
b. Write a formula for ππ in factored form using ππ for the
constant factor.
ππ(π₯π₯) = ππ(π₯π₯ + 10)(π₯π₯ + 1)(π₯π₯ β 2)
c. Use the fact that ππ(βππ) = βππππ to find the constant factor ππ.
β54 = ππ(β4 + 10)(β4 + 1)(β4β 2)
ππ = β12
ππ(π₯π₯) = β12
(π₯π₯ + 10)(π₯π₯ + 1)(π₯π₯ β 2)
d. Verify your equation by using the fact that ππ(ππ) = ππππ.
ππ(1) = β12
(1 + 10)(1 + 1)(1 β 2) = β12
(11)(2)(β1) = 11
Lesson 21: The Remainder Theorem
305
This work is derived from Eureka Math β’ and licensed by Great Minds. Β©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.
Hart Interactive β Algebra 2 M1 Lesson 21 ALGEBRA II
T
12. Find the value of ππ so that ππππβππππππ+ππππβππ
has remainder ππ.
ππ = β5
13. Find the value ππ so that ππππππ+ππβππππ+ππ
has remainder ππππ.
ππ = β2
14. Show that ππππππ β ππππππ + ππππ is divisible by ππ β ππ.
Let ππ(π₯π₯) = π₯π₯51 β 21π₯π₯ + 20.
Then ππ(1) = 151 β 21(1) + 20 = 1 β 21 + 20 = 0.
Since ππ(1) = 0, the remainder of the quotient (π₯π₯51 β 21π₯π₯ + 20) Γ· (π₯π₯ β 1) is 0.
Therefore, π₯π₯51 β 21π₯π₯ + 20 is divisible by π₯π₯ β 1.
15. Show that ππ + ππ is a factor of ππππππππππ + ππππππ β ππ.
Let ππ(π₯π₯) = 19π₯π₯42 + 18π₯π₯ β 1.
Then ππ(β1) = 19(β1)42 + 18(β1)β 1 = 19 β 18 β 1 = 0.
Since ππ(β1) = 0, π₯π₯ + 1 must be a factor of ππ.
Lesson 21: The Remainder Theorem
306
This work is derived from Eureka Math β’ and licensed by Great Minds. Β©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.