lesson 8: adding and subtracting polynomials

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M1 Lesson 8 ALGEBRA I Lesson 8: Adding and Subtracting Polynomials Student Outcomes Students understand that the sum or difference of two polynomials produces another polynomial and relate polynomials to the system of integers; students add and subtract polynomials. Classwork Exercise 1 (7 minutes) Have students complete Exercise 1(a) and use it for a brief discussion on the notion of base. Then have students continue with the remainder of the exercise. Exercise 1 a. How many quarters, nickels, and pennies are needed to make $. ? Answers will vary. quarters, nickels, pennies b. Fill in the blanks: , = Γ— + Γ— + Γ— + Γ— = Γ— + Γ— + Γ— + Γ— c. Fill in the blanks: , = Γ— + Γ— + Γ— + Γ— d. Fill in the blanks: = Γ— + Γ— + Γ— Next ask: Why do we use base 10? Why do we humans have a predilection for the number 10? Why do some cultures have base 20? How do you say β€œ87” in French? How does the Gettysburg address begin? Quatre-vingt-sept: 4-20s and 7; Four score and seven years ago… Computers use which base system? Base 2 Extension: Mayan, Aztec, and Celtic all used base 20. The word β€œscore” (which means 20) originated from the Celtic language. Students could be asked to research more on this and on the cultures who use or used base 5 and base 60. Lesson 8: Adding and Subtracting Polynomials 85 Β© 2014 Common Core, Inc. All rights reserved. commoncore.org A STORY OF FUNCTIONS

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Page 1: Lesson 8: Adding and Subtracting Polynomials

M1 Lesson 8 ALGEBRA I

Lesson 8: Adding and Subtracting Polynomials

Student Outcomes

Students understand that the sum or difference of two polynomials produces another polynomial and relate polynomials to the system of integers; students add and subtract polynomials.

Classwork

Exercise 1 (7 minutes)

Have students complete Exercise 1(a) and use it for a brief discussion on the notion of base. Then have students continue with the remainder of the exercise.

Exercise 1

a. How many quarters, nickels, and pennies are needed to make $𝟏𝟏.𝟏𝟏𝟏𝟏?

Answers will vary.

πŸ’πŸ’ quarters, 𝟐𝟐 nickels, 𝟏𝟏 pennies

b. Fill in the blanks:

πŸ–πŸ–,πŸ—πŸ—πŸ’πŸ’πŸπŸ = πŸ–πŸ– Γ— 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏+ πŸ—πŸ— Γ— 𝟏𝟏𝟏𝟏𝟏𝟏+ πŸ’πŸ’ Γ— 𝟏𝟏𝟏𝟏 + 𝟏𝟏 Γ— 𝟏𝟏 = πŸ–πŸ– Γ— 𝟏𝟏𝟏𝟏𝟏𝟏 + πŸ—πŸ— Γ— 𝟏𝟏𝟏𝟏𝟐𝟐 + πŸ’πŸ’ Γ— 𝟏𝟏𝟏𝟏+ 𝟏𝟏 Γ— 𝟏𝟏

c. Fill in the blanks:

πŸ–πŸ–,πŸ—πŸ—πŸ’πŸ’πŸπŸ = 𝟏𝟏 Γ— 𝟐𝟐𝟏𝟏𝟏𝟏 + 𝟐𝟐 Γ— 𝟐𝟐𝟏𝟏𝟐𝟐 + πŸ•πŸ• Γ— 𝟐𝟐𝟏𝟏 + 𝟏𝟏 Γ— 𝟏𝟏

d. Fill in the blanks:

𝟏𝟏𝟏𝟏𝟏𝟏 = πŸ’πŸ’ Γ— πŸ“πŸ“πŸπŸ + 𝟐𝟐 Γ— πŸ“πŸ“ + 𝟏𝟏 Γ— 𝟏𝟏

Next ask:

Why do we use base 10? Why do we humans have a predilection for the number 10?

Why do some cultures have base 20?

How do you say β€œ87” in French? How does the Gettysburg address begin?

Quatre-vingt-sept: 4-20s and 7; Four score and seven years ago…

Computers use which base system? Base 2

Extension: Mayan, Aztec, and

Celtic all used base 20. The word β€œscore” (which means 20) originated from the Celtic language.

Students could be asked to research more on this and on the cultures who use or used base 5 and base 60.

Lesson 8: Adding and Subtracting Polynomials

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Exercise 2 (5 minutes)

In Exercise 2, we are laying the foundation that polynomials written in standard form are simply base π‘₯π‘₯ β€œnumbers.” The practice of filling in specific values for π‘₯π‘₯ and finding the resulting values lays a foundation for connecting this algebra of polynomial expressions with the later lessons on polynomial functions (and other functions) and their inputs and outputs.

Work through Exercise 2 with the class.

Exercise 2

Now let’s be as general as possible by not identifying which base we are in. Just call the base 𝒙𝒙.

Consider the expression 𝟏𝟏 Γ— π’™π’™πŸπŸ + 𝟐𝟐 Γ— π’™π’™πŸπŸ + πŸ•πŸ• Γ— 𝒙𝒙 + 𝟏𝟏 Γ— 𝟏𝟏, or equivalently π’™π’™πŸπŸ + πŸπŸπ’™π’™πŸπŸ + πŸ•πŸ•π’™π’™ + 𝟏𝟏.

a. What is the value of this expression if 𝒙𝒙 = 𝟏𝟏𝟏𝟏?

𝟏𝟏,πŸπŸπŸ•πŸ•πŸπŸ

b. What is the value of this expression if 𝒙𝒙 = 𝟐𝟐𝟏𝟏?

πŸ–πŸ–,πŸ—πŸ—πŸ’πŸ’πŸπŸ

Point out that the expression we see here is just the generalized form of their answer from part (b) of Exercise 1. However, as we change π‘₯π‘₯, we get a different number each time.

Exercise 3 (10 minutes)

Allow students time to complete Exercise 3 individually. Then elicit responses from the class.

Exercise 3

a. When writing numbers in base 𝟏𝟏𝟏𝟏, we only allow coefficients of 𝟏𝟏 through πŸ—πŸ—. Why is that?

Once you get ten of a given unit, you also have one of the unit to the left of that.

b. What is the value of πŸπŸπŸπŸπ’™π’™ + 𝟏𝟏 when 𝒙𝒙 = πŸ“πŸ“? How much money is 𝟐𝟐𝟐𝟐 nickels and 𝟏𝟏 pennies?

𝟏𝟏𝟏𝟏𝟏𝟏

$𝟏𝟏.𝟏𝟏𝟏𝟏

c. What number is represented by πŸ’πŸ’π’™π’™πŸπŸ + πŸπŸπŸ•πŸ•π’™π’™ + 𝟐𝟐 if 𝒙𝒙 = 𝟏𝟏𝟏𝟏?

πŸ“πŸ“πŸ•πŸ•πŸπŸ

d. What number is represented by πŸ’πŸ’π’™π’™πŸπŸ + πŸπŸπŸ•πŸ•π’™π’™ + 𝟐𝟐 if 𝒙𝒙 = βˆ’πŸπŸ or if 𝒙𝒙 = 𝟐𝟐𝟏𝟏 ?

βˆ’πŸπŸπŸπŸ

πŸπŸπŸπŸπŸπŸπŸ—πŸ—

e. What number is represented by βˆ’πŸπŸπ’™π’™πŸπŸ + βˆšπŸπŸπ’™π’™ + 𝟏𝟏𝟐𝟐 when 𝒙𝒙 = √𝟐𝟐?

βˆ’πŸ•πŸ•πŸπŸ

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Point out, as highlighted by Exercises 1 and 3, that carrying is not necessary in this type of expression (polynomial expressions). For example, 4π‘₯π‘₯2 + 17π‘₯π‘₯ + 2 is a valid expression. However, in base ten arithmetic, coefficients of value ten or greater are not conventional notation. Setting π‘₯π‘₯ = 10 in 4π‘₯π‘₯2 + 17π‘₯π‘₯ + 2 yields 4 hundreds, 17 tens, and 2 ones, which is to be expressed as 5 hundreds, 7 tens, and 2 ones.

Discussion (11 minutes)

The next item in your student materials is a definition for a polynomial expression. Read the definition carefully, and then create 3 polynomial expressions using the given definition.

Polynomial Expression: A polynomial expression is either

1. A numerical expression or a variable symbol, or

2. The result of placing two previously generated polynomial expressions into the blanks of the addition operator (__+__) or the multiplication operator (__Γ—__).

Compare your polynomial expressions with a neighbor’s. Do your neighbor’s expressions fall into the category of polynomial expressions?

Resolve any debates as to whether a given expression is indeed a polynomial expression by referring back to the definition and discussing as a class.

Note that the definition of a polynomial expression includes subtraction (add the additive inverse instead), dividing by a non-zero number (multiply by the multiplicative inverse instead), and even exponentiation by a non-negative integer (use the multiplication operator repeatedly on the same numerical or variable symbol).

List several of the student-generated polynomials on the board. Include some that contain more than one variable.

Initiate the following discussion, presenting expressions on the board when relevant.

Just as the expression (3 + 4) βˆ™ 5 is a numerical expression but not a number, (π‘₯π‘₯ + 5) + (2π‘₯π‘₯2 βˆ’ π‘₯π‘₯)(3π‘₯π‘₯ + 1) is a polynomial expression but not technically a polynomial. We reserve the word polynomial for polynomial expressions that are written simply as a sum of monomial terms. This begs the question: What is a monomial?

A monomial is a polynomial expression generated using only the multiplication operator (__Γ—__). Thus, it does not contain + or βˆ’ operators.

Just as we would not typically write a number in factored form and still refer to it as a number (we might call it a number in factored form), similarly, we do not write a monomial in factored form and still refer to it as a monomial. We multiply any numerical factors together and condense multiple instances of a variable factor using (whole number) exponents.

Try creating a monomial.

Compare the monomial you created with your neighbor’s. Is your neighbor’s expression really a monomial? Is it written in the standard form we use for monomials?

There are also such things as binomials and trinomials. Can anyone make a conjecture about what a binomial is and what a trinomial is and how they are the same or different from a polynomial?

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Students may conjecture that a binomial has two of something and that a trinomial three of something. Further, they might conjecture that a polynomial has many of something. Allow for discussion and then state the following:

A binomial is the sum (or difference) of two monomials. A trinomial is the sum (or difference) of three monomials. A polynomial, as stated earlier, is the sum of one or more monomials.

The degree of a monomial is the sum of the exponents of the variable symbols that appear in the monomial.

The degree of a polynomial is the degree of the monomial term with the highest degree.

While polynomials can contain multiple variable symbols, most of our work with polynomials will be with polynomials in one variable.

What do polynomial expressions in one variable look like? Create a polynomial expression in one variable and compare with your neighbor.

Post some of the student generated polynomials in one variable on the board.

Let’s relate polynomials to the work we did at the beginning of the lesson.

Is this expression an integer in base 10? 10(100 + 22 βˆ’ 2) + 3(10) + 8 βˆ’ 2(2)

Is the expression equivalent to the integer 1,234? How did we find out?

We rewrote the first expression in our standard form, right?

Polynomials in one variable have a standard form as well. Use your intuition of what standard form of a polynomial might be to write this polynomial expression as a polynomial in standard form: 2π‘₯π‘₯(π‘₯π‘₯2 βˆ’ 3π‘₯π‘₯ + 1) βˆ’ (π‘₯π‘₯3 + 2) and compare your result with your neighbor.

Students should arrive at the answer π‘₯π‘₯3 βˆ’ 6π‘₯π‘₯2 + 2π‘₯π‘₯ βˆ’ 2.

Confirm that in standard form, we start with the highest degreed monomial, and continue in descending order.

The leading term of a polynomial is the term of highest degree that would be written first if the polynomial is put into standard form. The leading coefficient is the coefficient of the leading term.

What would you imagine we mean when we refer to the constant term of the polynomial?

A constant term is any term with no variables. To find β€œthe constant” term of a polynomial, be sure you have combined any and all constant terms into one single numerical term, written last if the polynomial is put into standard form. Note that a polynomial does not have to have a constant term (or could be said to have a constant term of 0).

As an extension for advanced students, assign the task of writing of a formal definition for standard form of a polynomial. The formal definition is provided below for your reference:

A polynomial expression with one variable symbol π‘₯π‘₯ is in standard form if it is expressed as, π‘Žπ‘Žπ‘›π‘›π‘₯π‘₯𝑛𝑛 +π‘Žπ‘Žπ‘›π‘›βˆ’1π‘₯π‘₯π‘›π‘›βˆ’1 + β‹―+ π‘Žπ‘Ž1π‘₯π‘₯ + π‘Žπ‘Ž0, where 𝑛𝑛 is a non-negative integer, and π‘Žπ‘Ž0, π‘Žπ‘Ž1, π‘Žπ‘Ž2, … , π‘Žπ‘Žπ‘›π‘› are constant coefficients with π‘Žπ‘Žπ‘›π‘› β‰  0. A polynomial expression in π‘₯π‘₯ that is in standard form is often called a polynomial in π‘₯π‘₯.

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Exercise 4 (5 minutes) Exercise 4

Find each sum or difference by combining the parts that are alike.

a. πŸ’πŸ’πŸπŸπŸ•πŸ•+ 𝟐𝟐𝟏𝟏𝟏𝟏 = πŸ’πŸ’ hundreds + 𝟏𝟏 tens + πŸ•πŸ• ones + 𝟐𝟐 hundreds + 𝟏𝟏 tens + 𝟏𝟏 ones

= 𝟏𝟏 hundreds + πŸ’πŸ’ tens + πŸ–πŸ– ones

b. (πŸ’πŸ’π’™π’™πŸπŸ + 𝒙𝒙 + πŸ•πŸ•) + (πŸπŸπ’™π’™πŸπŸ + πŸπŸπ’™π’™ + 𝟏𝟏)

πŸπŸπ’™π’™πŸπŸ + πŸ’πŸ’π’™π’™ + πŸ–πŸ–

c. (πŸπŸπ’™π’™πŸπŸ βˆ’ π’™π’™πŸπŸ + πŸ–πŸ–) βˆ’ (π’™π’™πŸπŸ + πŸ“πŸ“π’™π’™πŸπŸ + πŸ’πŸ’π’™π’™βˆ’ πŸ•πŸ•)

πŸπŸπ’™π’™πŸπŸ βˆ’ πŸπŸπ’™π’™πŸπŸ βˆ’ πŸ’πŸ’π’™π’™ + πŸπŸπŸ“πŸ“

d. 𝟏𝟏(π’™π’™πŸπŸ + πŸ–πŸ–π’™π’™) βˆ’ 𝟐𝟐(π’™π’™πŸπŸ + 𝟏𝟏𝟐𝟐)

π’™π’™πŸπŸ + πŸπŸπŸ’πŸ’π’™π’™ βˆ’ πŸπŸπŸ’πŸ’

e. (πŸ“πŸ“ βˆ’ 𝒕𝒕 βˆ’ π’•π’•πŸπŸ) + (πŸ—πŸ—π’•π’• + π’•π’•πŸπŸ)

πŸ–πŸ–π’•π’•+ πŸ“πŸ“

f. (πŸπŸπ’‘π’‘ + 𝟏𝟏) + 𝟏𝟏(𝒑𝒑 βˆ’ πŸ–πŸ–) βˆ’ (𝒑𝒑 + 𝟐𝟐)

πŸ–πŸ–π’‘π’‘ βˆ’ πŸ’πŸ’πŸ—πŸ—

Closing (3 minutes)

How are polynomials analogous to integers?

While integers are in base 10, polynomials are in base π‘₯π‘₯.

If you add two polynomials together, is the result sure to be another polynomial? The difference of two polynomials?

Students will likely reply, β€œyes,” based on the few examples and their intuition. Are you sure? Can you think of an example where adding or subtracting two polynomials does not result in a

polynomial?

Students thinking about π‘₯π‘₯2 βˆ’ π‘₯π‘₯2 = 0 could suggest not. At this point, review the definition of a polynomial. Constant symbols are polynomials.

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Exit Ticket (4 minutes)

Lesson Summary

A monomial is a polynomial expression generated using only the multiplication operator (__Γ—__). Thus, it does not contain + or βˆ’ operators. Monomials are written with numerical factors multiplied together and variable or other symbols each occurring one time (using exponents to condense multiple instances of the same variable).

A polynomial is the sum (or difference) of monomials.

The degree of a monomial is the sum of the exponents of the variable symbols that appear in the monomial.

The degree of a polynomial is the degree of the monomial term with the highest degree.

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Name ___________________________________________________ Date____________________

Lesson 8: Adding and Subtracting Polynomials

Exit Ticket 1. Must the sum of three polynomials again be a polynomial?

2. Find (𝑀𝑀2 βˆ’ 𝑀𝑀 + 1) + (𝑀𝑀3 βˆ’ 2𝑀𝑀2 + 99).

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Exit Ticket Sample Solutions

1. Must the sum of three polynomials again be a polynomial?

Yes.

2. Find (π’˜π’˜πŸπŸ βˆ’π’˜π’˜ + 𝟏𝟏) + (π’˜π’˜πŸπŸ βˆ’ πŸπŸπ’˜π’˜πŸπŸ + πŸ—πŸ—πŸ—πŸ—).

π’˜π’˜πŸπŸ βˆ’π’˜π’˜πŸπŸ βˆ’π’˜π’˜ + 𝟏𝟏𝟏𝟏𝟏𝟏

Problem Set Sample Solutions

1. Celina says that each of the following expressions is actually a binomial in disguise:

i. πŸ“πŸ“π’‚π’‚π’‚π’‚π’‚π’‚ βˆ’ πŸπŸπ’‚π’‚πŸπŸ + πŸπŸπ’‚π’‚π’‚π’‚π’‚π’‚

ii. πŸ“πŸ“π’™π’™πŸπŸ βˆ™ πŸπŸπ’™π’™πŸπŸ βˆ’ πŸπŸπŸπŸπ’™π’™πŸ’πŸ’ + πŸπŸπ’™π’™πŸ“πŸ“ + πŸπŸπ’™π’™ βˆ™ (βˆ’πŸπŸ)π’™π’™πŸ’πŸ’

iii. (𝒕𝒕+ 𝟐𝟐)𝟐𝟐 βˆ’ πŸ’πŸ’π’•π’•

iv. πŸ“πŸ“(𝒂𝒂 βˆ’ 𝟏𝟏) βˆ’ 𝟏𝟏𝟏𝟏(𝒂𝒂 βˆ’ 𝟏𝟏) + 𝟏𝟏𝟏𝟏𝟏𝟏(π’‚π’‚βˆ’ 𝟏𝟏)

v. (𝟐𝟐𝟐𝟐𝟐𝟐 βˆ’ 𝟐𝟐𝟐𝟐𝟐𝟐)𝟐𝟐 βˆ’ (𝟐𝟐𝟐𝟐𝟐𝟐 βˆ’ 𝟐𝟐𝟐𝟐𝟐𝟐) βˆ™ 𝟐𝟐𝟐𝟐

For example, she sees that the expression in (i) is algebraically equivalent to πŸπŸπŸπŸπ’‚π’‚π’‚π’‚π’‚π’‚ βˆ’ πŸπŸπ’‚π’‚πŸπŸ, which is indeed a binomial. (She is happy to write this as πŸπŸπŸπŸπ’‚π’‚π’‚π’‚π’‚π’‚+ (βˆ’πŸπŸ)π’‚π’‚πŸπŸ, if you prefer.)

Is she right about the remaining four expressions?

She is right about the remaining four expressions. They all can be expressed as binomials.

2. Janie writes a polynomial expression using only one variable, 𝒙𝒙, with degree 𝟏𝟏. Max writes a polynomial expression using only one variable, 𝒙𝒙, with degree πŸ•πŸ•.

a. What can you determine about the degree of the sum of Janie’s and Max’s polynomials?

The degree would be πŸ•πŸ•.

b. What can you determine about the degree of the difference of Janie’s and Max’s polynomials?

The degree would be πŸ•πŸ•.

3. Suppose Janie writes a polynomial expression using only one variable, 𝒙𝒙, with degree of πŸ“πŸ“, and Max writes a polynomial expression using only one variable, 𝒙𝒙, with degree of πŸ“πŸ“.

a. What can you determine about the degree of the sum of Janie’s and Max’s polynomials?

The maximum degree could be πŸ“πŸ“, but it could also be anything less than that. For example, if Janie’s polynomial were π’™π’™πŸ“πŸ“ + πŸπŸπ’™π’™ βˆ’ 𝟏𝟏, and Max’s were βˆ’π’™π’™πŸ“πŸ“ + πŸπŸπ’™π’™πŸπŸ + 𝟏𝟏, the degree of the sum is only 𝟐𝟐.

b. What can you determine about the degree of the difference of Janie’s and Max’s polynomials?

The maximum degree could be πŸ“πŸ“, but it could also be anything less than that.

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M1 Lesson 8 ALGEBRA I

4. Find each sum or difference by combining the parts that are alike.

a. (πŸπŸπ’‘π’‘ + πŸ’πŸ’) + πŸ“πŸ“(𝒑𝒑 βˆ’ 𝟏𝟏) βˆ’ (𝒑𝒑+ πŸ•πŸ•)

πŸπŸπ’‘π’‘ βˆ’ πŸ–πŸ–

b. (πŸ•πŸ•π’™π’™πŸ’πŸ’ + πŸ—πŸ—π’™π’™) βˆ’ 𝟐𝟐(π’™π’™πŸ’πŸ’ + 𝟏𝟏𝟏𝟏)

πŸ“πŸ“π’™π’™πŸ’πŸ’ + πŸ—πŸ—π’™π’™ βˆ’ 𝟐𝟐𝟏𝟏

c. (𝟏𝟏 βˆ’ 𝒕𝒕 βˆ’ π’•π’•πŸ’πŸ’) + (πŸ—πŸ—π’•π’• + π’•π’•πŸ’πŸ’)

πŸ–πŸ–π’•π’•+ 𝟏𝟏

d. (πŸ“πŸ“ βˆ’ π’•π’•πŸπŸ) + 𝟏𝟏(π’•π’•πŸπŸ βˆ’ πŸ–πŸ–) βˆ’ (π’•π’•πŸπŸ + 𝟏𝟏𝟐𝟐)

πŸ’πŸ’π’•π’•πŸπŸ βˆ’ πŸ“πŸ“πŸ“πŸ“

e. (πŸ–πŸ–π’™π’™πŸπŸ + πŸ“πŸ“π’™π’™) βˆ’ 𝟏𝟏(π’™π’™πŸπŸ + 𝟐𝟐)

πŸ“πŸ“π’™π’™πŸπŸ + πŸ“πŸ“π’™π’™ βˆ’ 𝟏𝟏

f. (πŸπŸπŸπŸπ’™π’™ + 𝟏𝟏) + 𝟐𝟐(𝒙𝒙 βˆ’ πŸ’πŸ’) βˆ’ (𝒙𝒙 βˆ’ πŸπŸπŸ“πŸ“)

πŸπŸπŸπŸπ’™π’™ + πŸ–πŸ–

g. (πŸπŸπŸπŸπ’™π’™πŸπŸ + πŸ“πŸ“π’™π’™) βˆ’ 𝟐𝟐(π’™π’™πŸπŸ + 𝟏𝟏)

πŸπŸπŸπŸπ’™π’™πŸπŸ + πŸ“πŸ“π’™π’™ βˆ’ 𝟐𝟐

h. (πŸ—πŸ— βˆ’ 𝒕𝒕 βˆ’ π’•π’•πŸπŸ) βˆ’ 𝟏𝟏𝟐𝟐 (πŸ–πŸ–π’•π’•+ πŸπŸπ’•π’•πŸπŸ)

βˆ’πŸ’πŸ’π’•π’•πŸπŸ βˆ’ πŸπŸπŸπŸπ’•π’• + πŸ—πŸ—

i. (πŸ’πŸ’π’Žπ’Ž+ 𝟏𝟏) βˆ’ 𝟏𝟏𝟐𝟐(π’Žπ’Žβˆ’πŸπŸ) + (π’Žπ’Ž + 𝟐𝟐)

βˆ’πŸ•πŸ•π’Žπ’Ž+ πŸ’πŸ’πŸ’πŸ’

j. (πŸπŸπŸ“πŸ“π’™π’™πŸ’πŸ’ + πŸπŸπŸπŸπ’™π’™) βˆ’ 𝟏𝟏𝟐𝟐(π’™π’™πŸ’πŸ’ + πŸ’πŸ’π’™π’™)

πŸπŸπ’™π’™πŸ’πŸ’ βˆ’ πŸπŸπŸ–πŸ–π’™π’™

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M1 Lesson 9 ALGEBRA I

Lesson 9: Multiplying Polynomials

Student Outcomes

Students understand that the product of two polynomials produces another polynomial; students multiply polynomials.

Classwork

Exercise 1 (15 minutes) Exercise 1

a. Gisella computed πŸ‘πŸ‘πŸ‘πŸ‘πŸ‘πŸ‘Γ— πŸ‘πŸ‘πŸ‘πŸ‘ as follows:

Can you explain what she is doing? What is her final answer?

She is using an area model, finding the area of each rectangle and adding them together. Her final answer is πŸ•πŸ•,πŸ–πŸ–πŸ–πŸ–πŸ–πŸ–.

Use a geometric diagram to compute the following products:

b. (πŸ‘πŸ‘π’™π’™πŸ‘πŸ‘ + πŸ‘πŸ‘π’™π’™ + πŸ‘πŸ‘) Γ— (πŸ‘πŸ‘π’™π’™ + πŸ‘πŸ‘)

πŸ–πŸ–π’™π’™πŸ‘πŸ‘ + πŸπŸπŸ•πŸ•π’™π’™πŸ‘πŸ‘ + πŸπŸπŸ–πŸ–π’™π’™ + πŸ–πŸ–

c. (πŸ‘πŸ‘π’™π’™πŸ‘πŸ‘ + πŸπŸπŸπŸπ’™π’™ + 𝟏𝟏)(π’™π’™πŸ‘πŸ‘ + 𝒙𝒙 + 𝟏𝟏)

πŸ‘πŸ‘π’™π’™πŸ‘πŸ‘ + πŸπŸπŸ‘πŸ‘π’™π’™πŸ‘πŸ‘ + πŸπŸπŸ‘πŸ‘π’™π’™πŸ‘πŸ‘ + πŸπŸπŸπŸπ’™π’™ + 𝟏𝟏

d. (𝒙𝒙 βˆ’ 𝟏𝟏)(π’™π’™πŸ‘πŸ‘ + πŸ–πŸ–π’™π’™πŸ‘πŸ‘ βˆ’ πŸ“πŸ“)

π’™π’™πŸ‘πŸ‘ + πŸ“πŸ“π’™π’™πŸ‘πŸ‘ βˆ’ πŸ–πŸ–π’™π’™πŸ‘πŸ‘ βˆ’ πŸ“πŸ“π’™π’™ + πŸ“πŸ“

Ask the students:

What do you notice about the terms along the diagonals in the rectangles you drew?

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Encourage students to recognize that in parts (b) and (c), the terms along the diagonals were all like terms; however, in part (d) one of the factors has no π‘₯π‘₯-term. Allow students to develop a strategy for dealing with this, concluding with the suggestion of inserting the term + 0π‘₯π‘₯, for a model that looks like the following:

Students may naturally ask about the division of polynomials. This topic will be covered in Grade 11, Module 1. The extension challenge at the end of the lesson, however, could be of interest to students inquiring about this.

Could we have found this product without the aid of a geometric model? What would that look like?

Go through the exercise applying the distributive property and collecting like terms. As a scaffold, remind students that variables are placeholders for numbers. If π‘₯π‘₯ = 5, for example, whatever the quantity on the right is (270), you have 5 βˆ’ 1 of β€œthat quantity”, or 5 of β€œthat quantity” minus 1 of β€œthat quantity”. Similarly we have π‘₯π‘₯ of that quantity, minus 1 of that quantity:

(π‘₯π‘₯ βˆ’ 1)(π‘₯π‘₯3 + 6π‘₯π‘₯2 βˆ’ 5)

π‘₯π‘₯(π‘₯π‘₯3 + 6π‘₯π‘₯2 βˆ’ 5) βˆ’ 1(π‘₯π‘₯3 + 6π‘₯π‘₯2 βˆ’ 5)

π‘₯π‘₯4 + 6π‘₯π‘₯3 βˆ’ 5π‘₯π‘₯ βˆ’ π‘₯π‘₯3 βˆ’ 6π‘₯π‘₯2 + 5

π‘₯π‘₯4 + 5π‘₯π‘₯3 βˆ’ 6π‘₯π‘₯2 βˆ’ 5π‘₯π‘₯ + 5

Exercise 2 (5 minutes)

Have students work Exercise 2 independently and then compare answers with a neighbor. If needed, facilitate agreement on the correct answer by allowing students to discuss their thought processes and justify their solutions.

Exercise 2

Multiply the polynomials using the distributive property: (πŸ‘πŸ‘π’™π’™πŸ‘πŸ‘ + 𝒙𝒙 βˆ’ 𝟏𝟏)(π’™π’™πŸ‘πŸ‘ βˆ’ πŸ‘πŸ‘π’™π’™ + 𝟏𝟏).

πŸ‘πŸ‘π’™π’™πŸ‘πŸ‘(π’™π’™πŸ‘πŸ‘ βˆ’ πŸ‘πŸ‘π’™π’™ + 𝟏𝟏) + 𝒙𝒙(π’™π’™πŸ‘πŸ‘ βˆ’ πŸ‘πŸ‘π’™π’™ + 𝟏𝟏) βˆ’ 𝟏𝟏(π’™π’™πŸ‘πŸ‘ βˆ’ πŸ‘πŸ‘π’™π’™ + 𝟏𝟏)

πŸ‘πŸ‘π’™π’™πŸ–πŸ– βˆ’ πŸ–πŸ–π’™π’™πŸ‘πŸ‘ + πŸ‘πŸ‘π’™π’™πŸ‘πŸ‘ + π’™π’™πŸ“πŸ“ βˆ’ πŸ‘πŸ‘π’™π’™πŸ‘πŸ‘ + 𝒙𝒙 βˆ’ π’™π’™πŸ‘πŸ‘ + πŸ‘πŸ‘π’™π’™ βˆ’ 𝟏𝟏

πŸ‘πŸ‘π’™π’™πŸ–πŸ– + π’™π’™πŸ“πŸ“ βˆ’ π’™π’™πŸ‘πŸ‘ βˆ’ πŸ–πŸ–π’™π’™πŸ‘πŸ‘ + π’™π’™πŸ‘πŸ‘ + πŸ‘πŸ‘π’™π’™ βˆ’ 𝟏𝟏

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Exercise 3 (10 minutes)

Give students 10 minutes to complete Exercise 3 and compare their answers with a neighbor.

Exercise 3

The expression πŸπŸπŸπŸπ’™π’™πŸ‘πŸ‘ + πŸ–πŸ–π’™π’™πŸ‘πŸ‘ is the result of applying the distributive property to the expression πŸ‘πŸ‘π’™π’™πŸ‘πŸ‘(πŸ“πŸ“+ πŸ‘πŸ‘π’™π’™). It is also the result of the applying the distributive property to πŸ‘πŸ‘(πŸ“πŸ“π’™π’™πŸ‘πŸ‘ + πŸ‘πŸ‘π’™π’™πŸ‘πŸ‘) or to 𝒙𝒙(πŸπŸπŸπŸπ’™π’™ + πŸ–πŸ–π’™π’™πŸ‘πŸ‘), for example, or even to 𝟏𝟏 βˆ™(πŸπŸπŸπŸπ’™π’™πŸ‘πŸ‘ + πŸ–πŸ–π’™π’™πŸ‘πŸ‘)!

For (i) to (x) below, write down an expression such that if you applied the distributive property to your expression it will give the result presented. Give interesting answers!

i. πŸ–πŸ–π’‚π’‚ + πŸπŸπŸ‘πŸ‘π’‚π’‚πŸ‘πŸ‘

ii. πŸ‘πŸ‘π’™π’™πŸ‘πŸ‘ + πŸ‘πŸ‘π’™π’™πŸ“πŸ“ + πŸ‘πŸ‘π’™π’™πŸπŸπŸπŸ

iii. πŸ–πŸ–π’›π’›πŸ‘πŸ‘ βˆ’ πŸπŸπŸ“πŸ“π’›π’›

iv. πŸ‘πŸ‘πŸ‘πŸ‘π’˜π’˜πŸ‘πŸ‘ βˆ’ πŸπŸπŸ‘πŸ‘π’˜π’˜ + πŸ•πŸ•πŸ•πŸ•π’˜π’˜πŸ“πŸ“

v. π’›π’›πŸ‘πŸ‘(𝒂𝒂 + 𝒃𝒃) + π’›π’›πŸ‘πŸ‘(𝒂𝒂+ 𝒃𝒃)

vi. πŸ‘πŸ‘πŸ‘πŸ‘π’”π’”πŸ‘πŸ‘ +

πŸπŸπŸ‘πŸ‘

vii. πŸπŸπŸ“πŸ“π’‘π’‘πŸ‘πŸ‘π’“π’“πŸ‘πŸ‘ βˆ’ πŸ–πŸ–π’‘π’‘πŸ‘πŸ‘π’“π’“πŸ“πŸ“ + πŸ—πŸ—π’‘π’‘πŸ‘πŸ‘π’“π’“πŸ‘πŸ‘ + πŸ‘πŸ‘βˆšπŸ‘πŸ‘π’‘π’‘πŸ‘πŸ‘π’“π’“πŸ–πŸ–

viii. 𝟏𝟏.πŸ‘πŸ‘π’™π’™πŸ—πŸ— βˆ’ πŸ‘πŸ‘πŸπŸπ’™π’™πŸ–πŸ–

ix. (πŸ‘πŸ‘π’™π’™ + πŸ‘πŸ‘)(π’™π’™πŸ‘πŸ‘ + π’™π’™πŸ‘πŸ‘) βˆ’ (πŸ‘πŸ‘π’™π’™ + πŸ‘πŸ‘)(π’™π’™πŸ‘πŸ‘ + π’™π’™πŸ‘πŸ‘)

x. (πŸ‘πŸ‘π’›π’› + πŸ“πŸ“)(𝒛𝒛 βˆ’ πŸ‘πŸ‘) βˆ’ (πŸπŸπŸ‘πŸ‘π’›π’› βˆ’ πŸ‘πŸ‘πŸ–πŸ–)(𝒛𝒛 βˆ’ πŸ‘πŸ‘)

Some possible answers:

i. πŸ‘πŸ‘π’‚π’‚(πŸ‘πŸ‘+ πŸ•πŸ•π’‚π’‚) or πŸ‘πŸ‘(πŸ‘πŸ‘π’‚π’‚ + πŸ•πŸ•π’‚π’‚πŸ‘πŸ‘) or 𝒂𝒂(πŸ–πŸ–+ πŸπŸπŸ‘πŸ‘π’‚π’‚)

ii. πŸ‘πŸ‘π’™π’™πŸ‘πŸ‘(𝟏𝟏+ 𝒙𝒙 + π’™π’™πŸ–πŸ–) or 𝒙𝒙(πŸ‘πŸ‘π’™π’™πŸ‘πŸ‘ + πŸ‘πŸ‘π’™π’™πŸ‘πŸ‘ + πŸ‘πŸ‘π’™π’™πŸ—πŸ—) or πŸ‘πŸ‘(π’™π’™πŸ‘πŸ‘ + π’™π’™πŸ“πŸ“ + π’™π’™πŸπŸπŸπŸ)

iii. πŸ‘πŸ‘π’›π’›(πŸ‘πŸ‘π’›π’› βˆ’ πŸ“πŸ“) or πŸ‘πŸ‘(πŸ‘πŸ‘π’›π’›πŸ‘πŸ‘ βˆ’ πŸ“πŸ“π’›π’›) or 𝒛𝒛(πŸ–πŸ–π’›π’› βˆ’ πŸπŸπŸ“πŸ“)

iv. πŸ•πŸ•π’˜π’˜(πŸ–πŸ–π’˜π’˜πŸ‘πŸ‘ βˆ’ πŸ‘πŸ‘ + πŸπŸπŸπŸπ’˜π’˜πŸ‘πŸ‘) or π’˜π’˜(πŸ‘πŸ‘πŸ‘πŸ‘π’˜π’˜πŸ‘πŸ‘ βˆ’ πŸπŸπŸ‘πŸ‘ + πŸ•πŸ•πŸ•πŸ•π’˜π’˜πŸ‘πŸ‘)

v. π’›π’›πŸ‘πŸ‘οΏ½(𝒂𝒂+ 𝒃𝒃) + 𝒛𝒛(𝒂𝒂+ 𝒃𝒃)οΏ½ or 𝒛𝒛(𝒛𝒛(𝒂𝒂 + 𝒃𝒃) + π’›π’›πŸ‘πŸ‘(𝒂𝒂+ 𝒃𝒃))

vi. πŸπŸπŸ‘πŸ‘

(πŸ‘πŸ‘π’”π’”πŸ‘πŸ‘ + 𝟏𝟏)

vii. πŸ‘πŸ‘π’‘π’‘πŸ‘πŸ‘π’“π’“πŸ‘πŸ‘οΏ½πŸ“πŸ“π’‘π’‘π’“π’“πŸ‘πŸ‘ βˆ’ πŸ‘πŸ‘π’“π’“πŸ‘πŸ‘ + πŸ‘πŸ‘π’‘π’‘πŸ‘πŸ‘ + βˆšπŸ‘πŸ‘π’‘π’‘π’“π’“πŸ‘πŸ‘οΏ½ or π’‘π’‘πŸ‘πŸ‘π’“π’“πŸ‘πŸ‘(πŸπŸπŸ“πŸ“π’‘π’‘π’“π’“πŸ‘πŸ‘ βˆ’ πŸ–πŸ–π’“π’“πŸ‘πŸ‘ + πŸ—πŸ—π’‘π’‘πŸ‘πŸ‘ + πŸ‘πŸ‘βˆšπŸ‘πŸ‘π’‘π’‘π’“π’“πŸ‘πŸ‘)

viii. 𝟏𝟏.πŸ‘πŸ‘π’™π’™πŸ–πŸ–(𝒙𝒙 βˆ’ 𝟏𝟏𝟏𝟏𝟏𝟏) or πŸ‘πŸ‘πŸπŸπŸπŸ π’™π’™πŸ–πŸ–(𝒙𝒙 βˆ’ 𝟏𝟏𝟏𝟏𝟏𝟏)

ix. (π’™π’™πŸ‘πŸ‘ + π’™π’™πŸ‘πŸ‘)((πŸ‘πŸ‘π’™π’™+ πŸ‘πŸ‘) βˆ’ (πŸ‘πŸ‘π’™π’™ + πŸ‘πŸ‘))

x. (𝒛𝒛 βˆ’ πŸ‘πŸ‘)οΏ½(πŸ‘πŸ‘π’›π’›+ πŸ“πŸ“) βˆ’ πŸπŸπŸ‘πŸ‘(𝒛𝒛 βˆ’ πŸ‘πŸ‘)οΏ½

Choose one (or more) to go through as a class, listing as many different re-writes as possible. Then remark:

The process of making use of the distributive property β€œbackwards” is factoring.

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Exercise 4 (5 minutes) Exercise 4

Sammy wrote a polynomial using only one variable, 𝒙𝒙, of degree πŸ‘πŸ‘. Myisha wrote a polynomial in the same variable of degree πŸ“πŸ“. What can you say about the degree of the product of Sammy’s and Myisha’s polynomials?

The degree of the product of the two polynomials would be πŸ–πŸ–.

Extension Extension

Find a polynomial that, when multiplied by πŸ‘πŸ‘π’™π’™πŸ‘πŸ‘ + πŸ‘πŸ‘π’™π’™ + 𝟏𝟏, gives the answer πŸ‘πŸ‘π’™π’™πŸ‘πŸ‘ + π’™π’™πŸ‘πŸ‘ βˆ’ πŸ‘πŸ‘π’™π’™ βˆ’ 𝟏𝟏.

𝒙𝒙 βˆ’ 𝟏𝟏

Closing (6 minutes)

Is the product of two polynomials sure to be another polynomial?

Yes, by the definition of polynomial expression given in Lesson 8, the product of any two polynomial expressions is again a polynomial expression, which can then be written in standard polynomial form through application of the distributive property.

Is a polynomial squared sure to be another polynomial (other integer powers)? Yes.

Exit Ticket (4 minutes)

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Name ___________________________________________________ Date____________________

Lesson 9: Multiplying Polynomials

Exit Ticket 1. Must the product of three polynomials again be a polynomial?

2. Find (𝑀𝑀2 + 1)(𝑀𝑀3 βˆ’ 𝑀𝑀 + 1).

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M1 Lesson 9 ALGEBRA I

Exit Ticket Sample Solutions

1. Must the product of three polynomials again be a polynomial?

Yes.

2. Find (π’˜π’˜πŸ‘πŸ‘ + 𝟏𝟏)(π’˜π’˜πŸ‘πŸ‘ βˆ’ π’˜π’˜ + 𝟏𝟏).

π’˜π’˜πŸ“πŸ“ +π’˜π’˜πŸ‘πŸ‘ βˆ’π’˜π’˜ + 𝟏𝟏

Problem Set Sample Solutions

1. Use the distributive property to write each of the following expressions as the sum of monomials.

a. πŸ‘πŸ‘π’‚π’‚(πŸ‘πŸ‘+ 𝒂𝒂)

πŸ‘πŸ‘π’‚π’‚πŸ‘πŸ‘ + πŸπŸπŸ‘πŸ‘π’‚π’‚

b. 𝒙𝒙(𝒙𝒙+ πŸ‘πŸ‘) + 𝟏𝟏

π’™π’™πŸ‘πŸ‘ + πŸ‘πŸ‘π’™π’™ + 𝟏𝟏

c. πŸπŸπŸ‘πŸ‘

(πŸπŸπŸ‘πŸ‘π’›π’› + πŸπŸπŸ–πŸ–π’›π’›πŸ‘πŸ‘)

πŸ–πŸ–π’›π’›πŸ‘πŸ‘ + πŸ‘πŸ‘π’›π’›

d. πŸ‘πŸ‘π’™π’™(π’™π’™πŸ‘πŸ‘ βˆ’ 𝟏𝟏𝟏𝟏)

πŸ‘πŸ‘π’™π’™πŸ‘πŸ‘ βˆ’ πŸ‘πŸ‘πŸπŸπ’™π’™

e. (𝒙𝒙 βˆ’ πŸ‘πŸ‘)(𝒙𝒙+ πŸ“πŸ“)

π’™π’™πŸ‘πŸ‘ + 𝒙𝒙 βˆ’ πŸ‘πŸ‘πŸπŸ

f. (πŸ‘πŸ‘π’›π’› βˆ’ 𝟏𝟏)(πŸ‘πŸ‘π’›π’›πŸ‘πŸ‘ + 𝟏𝟏)

πŸ–πŸ–π’›π’›πŸ‘πŸ‘ βˆ’ πŸ‘πŸ‘π’›π’›πŸ‘πŸ‘ + πŸ‘πŸ‘π’›π’› βˆ’ 𝟏𝟏

g. (πŸπŸπŸπŸπ’˜π’˜βˆ’ 𝟏𝟏)(πŸπŸπŸπŸπ’˜π’˜ + 𝟏𝟏)

πŸπŸπŸπŸπŸπŸπ’˜π’˜πŸ‘πŸ‘ βˆ’ 𝟏𝟏

h. (βˆ’πŸ“πŸ“π’˜π’˜βˆ’ πŸ‘πŸ‘)π’˜π’˜πŸ‘πŸ‘ βˆ’πŸ“πŸ“π’˜π’˜πŸ‘πŸ‘ βˆ’ πŸ‘πŸ‘π’˜π’˜πŸ‘πŸ‘

i. πŸπŸπŸ–πŸ–π’”π’”πŸπŸπŸπŸπŸπŸ οΏ½πŸπŸπŸ‘πŸ‘ π’”π’”πŸ‘πŸ‘πŸπŸπŸπŸ + 𝟏𝟏.πŸπŸπŸ‘πŸ‘πŸ“πŸ“π’”π’”οΏ½

πŸ–πŸ–π’”π’”πŸ‘πŸ‘πŸπŸπŸπŸ + πŸ‘πŸ‘π’”π’”πŸπŸπŸπŸπŸπŸ

j. (πŸ‘πŸ‘π’’π’’ + 𝟏𝟏)(πŸ‘πŸ‘π’’π’’πŸ‘πŸ‘ + 𝟏𝟏)

πŸ‘πŸ‘π’’π’’πŸ‘πŸ‘ + πŸ‘πŸ‘π’’π’’πŸ‘πŸ‘ + πŸ‘πŸ‘π’’π’’ + 𝟏𝟏

k. (π’™π’™πŸ‘πŸ‘ βˆ’ 𝒙𝒙 + 𝟏𝟏)(𝒙𝒙 βˆ’ 𝟏𝟏)

π’™π’™πŸ‘πŸ‘ βˆ’ πŸ‘πŸ‘π’™π’™πŸ‘πŸ‘ + πŸ‘πŸ‘π’™π’™ βˆ’ 𝟏𝟏

l. πŸ‘πŸ‘π’™π’™π’›π’›(πŸ—πŸ—π’™π’™π’™π’™+ 𝒛𝒛) βˆ’ πŸ‘πŸ‘π’™π’™π’›π’›(𝒙𝒙 + 𝒙𝒙 βˆ’ 𝒛𝒛)

πŸ‘πŸ‘πŸ•πŸ•π’™π’™πŸ‘πŸ‘π’™π’™π’›π’›+ πŸ‘πŸ‘π’™π’™π’›π’›πŸ‘πŸ‘ βˆ’ πŸ‘πŸ‘π’™π’™π’™π’™π’›π’› βˆ’ πŸ‘πŸ‘π’™π’™πŸ‘πŸ‘π’›π’›+ πŸ‘πŸ‘π’™π’™π’›π’›πŸ‘πŸ‘

m. (𝒕𝒕 βˆ’ 𝟏𝟏)(𝒕𝒕+ 𝟏𝟏)(π’•π’•πŸ‘πŸ‘ + 𝟏𝟏)

π’•π’•πŸ‘πŸ‘ βˆ’ 𝟏𝟏

n. (π’˜π’˜ + 𝟏𝟏)(π’˜π’˜πŸ‘πŸ‘ βˆ’ π’˜π’˜πŸ‘πŸ‘ +π’˜π’˜πŸ‘πŸ‘ βˆ’π’˜π’˜ + 𝟏𝟏)

π’˜π’˜πŸ“πŸ“ + 𝟏𝟏

o. 𝒛𝒛(πŸ‘πŸ‘π’›π’›+ 𝟏𝟏)(πŸ‘πŸ‘π’›π’› βˆ’ πŸ‘πŸ‘)

πŸ–πŸ–π’›π’›πŸ‘πŸ‘ βˆ’ π’›π’›πŸ‘πŸ‘ βˆ’ πŸ‘πŸ‘π’›π’›

p. (𝒙𝒙 + 𝒙𝒙)(𝒙𝒙+ 𝒛𝒛)(𝒛𝒛+ 𝒙𝒙)

πŸ‘πŸ‘π’™π’™π’™π’™π’›π’› + π’™π’™πŸ‘πŸ‘π’™π’™+ π’™π’™πŸ‘πŸ‘π’›π’›+ π’™π’™π’™π’™πŸ‘πŸ‘ + π’™π’™π’›π’›πŸ‘πŸ‘ + π’™π’™πŸ‘πŸ‘π’›π’› + π’™π’™π’›π’›πŸ‘πŸ‘

q. 𝒙𝒙+π’™π’™πŸ‘πŸ‘

πŸπŸπŸ‘πŸ‘π’™π’™ +

πŸπŸπŸ‘πŸ‘π’™π’™

r. (πŸ‘πŸ‘πŸπŸπ’‡π’‡πŸπŸπŸπŸ βˆ’ πŸπŸπŸπŸπ’‡π’‡πŸ“πŸ“) Γ· πŸ“πŸ“

πŸ‘πŸ‘π’‡π’‡πŸπŸπŸπŸ βˆ’ πŸ‘πŸ‘π’‡π’‡πŸ“πŸ“

s. βˆ’πŸ“πŸ“π’™π’™(π’™π’™πŸ‘πŸ‘ + 𝒙𝒙 βˆ’ πŸ‘πŸ‘) βˆ’ πŸ‘πŸ‘(πŸ‘πŸ‘ βˆ’ π’™π’™πŸ‘πŸ‘)

βˆ’πŸ‘πŸ‘π’™π’™πŸ‘πŸ‘ βˆ’ πŸ“πŸ“π’™π’™πŸ‘πŸ‘ + πŸπŸπŸπŸπ’™π’™ βˆ’ πŸ‘πŸ‘ t.

(𝒂𝒂+π’ƒπ’ƒβˆ’π’„π’„)(𝒂𝒂+𝒃𝒃+𝒄𝒄)πŸπŸπŸ•πŸ•

πŸπŸπŸπŸπŸ•πŸ•

π’‚π’‚πŸ‘πŸ‘ +πŸπŸπŸπŸπŸ•πŸ•

π’ƒπ’ƒπŸ‘πŸ‘ βˆ’πŸπŸπŸπŸπŸ•πŸ•

π’„π’„πŸ‘πŸ‘ +πŸ‘πŸ‘πŸπŸπŸ•πŸ•

𝒂𝒂𝒃𝒃

u. (πŸ‘πŸ‘π’™π’™ Γ· πŸ—πŸ— + (πŸ“πŸ“π’™π’™) Γ· πŸ‘πŸ‘) Γ· (βˆ’πŸ‘πŸ‘)

βˆ’πŸ‘πŸ‘πŸ—πŸ—π’™π’™πŸ‘πŸ‘πŸ–πŸ–

v. (βˆ’πŸ‘πŸ‘π’‡π’‡πŸ‘πŸ‘ βˆ’ πŸ‘πŸ‘π’‡π’‡ + 𝟏𝟏)(π’‡π’‡πŸ‘πŸ‘ βˆ’ 𝒇𝒇 + πŸ‘πŸ‘)

βˆ’πŸ‘πŸ‘π’‡π’‡πŸ“πŸ“ + πŸ‘πŸ‘π’‡π’‡πŸ‘πŸ‘ βˆ’ πŸ–πŸ–π’‡π’‡πŸ‘πŸ‘ + πŸ‘πŸ‘π’‡π’‡πŸ‘πŸ‘ βˆ’ πŸ“πŸ“π’‡π’‡ + πŸ‘πŸ‘

Lesson 9: Multiplying Polynomials

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M1 Lesson 9 ALGEBRA I

2. Use the distributive property (and your wits!) to write each of the following expressions as a sum of monomials. If the resulting polynomial is in one variable, write the polynomial in standard form.

a. (𝒂𝒂 + 𝒃𝒃)πŸ‘πŸ‘

π’‚π’‚πŸ‘πŸ‘ + πŸ‘πŸ‘π’‚π’‚π’ƒπ’ƒ + π’ƒπ’ƒπŸ‘πŸ‘

b. (𝒂𝒂 + 𝟏𝟏)πŸ‘πŸ‘

π’‚π’‚πŸ‘πŸ‘ + πŸ‘πŸ‘π’‚π’‚ + 𝟏𝟏

c. (πŸ‘πŸ‘+ 𝒃𝒃)πŸ‘πŸ‘

π’ƒπ’ƒπŸ‘πŸ‘ + πŸ–πŸ–π’ƒπ’ƒ + πŸ—πŸ—

d. (πŸ‘πŸ‘+ 𝟏𝟏)πŸ‘πŸ‘

πŸπŸπŸ–πŸ–

e. (𝒙𝒙 + 𝒙𝒙+ 𝒛𝒛)πŸ‘πŸ‘

π’™π’™πŸ‘πŸ‘ + π’™π’™πŸ‘πŸ‘ + π’›π’›πŸ‘πŸ‘ + πŸ‘πŸ‘π’™π’™π’™π’™+ πŸ‘πŸ‘π’™π’™π’›π’› + πŸ‘πŸ‘π’™π’™π’›π’›

f. (𝒙𝒙 + 𝟏𝟏+ 𝒛𝒛)πŸ‘πŸ‘

π’™π’™πŸ‘πŸ‘ + π’›π’›πŸ‘πŸ‘ + πŸ‘πŸ‘π’™π’™π’›π’›+ πŸ‘πŸ‘π’™π’™ + πŸ‘πŸ‘π’›π’› + 𝟏𝟏

g. (πŸ‘πŸ‘+ 𝒛𝒛)πŸ‘πŸ‘

π’›π’›πŸ‘πŸ‘ + πŸ–πŸ–π’›π’› + πŸ—πŸ—

h. (𝒑𝒑 + 𝒒𝒒)πŸ‘πŸ‘

π’‘π’‘πŸ‘πŸ‘ + πŸ‘πŸ‘π’‘π’‘πŸ‘πŸ‘π’’π’’ + πŸ‘πŸ‘π’‘π’‘π’’π’’πŸ‘πŸ‘ + π’’π’’πŸ‘πŸ‘

i. (𝒑𝒑 βˆ’ 𝟏𝟏)πŸ‘πŸ‘

π’‘π’‘πŸ‘πŸ‘ βˆ’ πŸ‘πŸ‘π’‘π’‘πŸ‘πŸ‘ + πŸ‘πŸ‘π’‘π’‘ βˆ’ 𝟏𝟏

j. (πŸ“πŸ“+ 𝒒𝒒)πŸ‘πŸ‘

π’’π’’πŸ‘πŸ‘ + πŸπŸπŸ“πŸ“π’’π’’πŸ‘πŸ‘ + πŸ•πŸ•πŸ“πŸ“π’’π’’+ πŸπŸπŸ‘πŸ‘πŸ“πŸ“

3. Use the distributive property (and your wits!) to write each of the following expressions as a polynomial in standard form.

a. (π’”π’”πŸ‘πŸ‘ + πŸ‘πŸ‘)(𝒔𝒔 βˆ’ 𝟏𝟏)

π’”π’”πŸ‘πŸ‘ βˆ’ π’”π’”πŸ‘πŸ‘ + πŸ‘πŸ‘π’”π’” βˆ’ πŸ‘πŸ‘

b. πŸ‘πŸ‘(π’”π’”πŸ‘πŸ‘ + πŸ‘πŸ‘)(𝒔𝒔 βˆ’ 𝟏𝟏)

πŸ‘πŸ‘π’”π’”πŸ‘πŸ‘ βˆ’ πŸ‘πŸ‘π’”π’”πŸ‘πŸ‘ + πŸπŸπŸ‘πŸ‘π’”π’” βˆ’ πŸπŸπŸ‘πŸ‘

c. 𝒔𝒔(π’”π’”πŸ‘πŸ‘ + πŸ‘πŸ‘)(𝒔𝒔 βˆ’ 𝟏𝟏)

π’”π’”πŸ‘πŸ‘ βˆ’ π’”π’”πŸ‘πŸ‘ + πŸ‘πŸ‘π’”π’”πŸ‘πŸ‘ βˆ’ πŸ‘πŸ‘π’”π’”

d. (𝒔𝒔 + 𝟏𝟏)(π’”π’”πŸ‘πŸ‘ + πŸ‘πŸ‘)(𝒔𝒔 βˆ’ 𝟏𝟏)

π’”π’”πŸ‘πŸ‘ + πŸ‘πŸ‘π’”π’”πŸ‘πŸ‘ βˆ’ πŸ‘πŸ‘

e. (𝒖𝒖 βˆ’ 𝟏𝟏)(π’–π’–πŸ“πŸ“ + π’–π’–πŸ‘πŸ‘ + π’–π’–πŸ‘πŸ‘ + π’–π’–πŸ‘πŸ‘ + 𝒖𝒖 + 𝟏𝟏)

π’–π’–πŸ–πŸ– βˆ’ 𝟏𝟏 f. βˆšπŸ“πŸ“(π’–π’–βˆ’ 𝟏𝟏)(π’–π’–πŸ“πŸ“ + π’–π’–πŸ‘πŸ‘ + π’–π’–πŸ‘πŸ‘ + π’–π’–πŸ‘πŸ‘ + 𝒖𝒖+ 𝟏𝟏)

βˆšπŸ“πŸ“π’–π’–πŸ–πŸ– βˆ’ βˆšπŸ“πŸ“

g. (π’–π’–πŸ•πŸ• + π’–π’–πŸ‘πŸ‘ + 𝟏𝟏)(π’–π’–βˆ’ 𝟏𝟏)(π’–π’–πŸ“πŸ“ + π’–π’–πŸ‘πŸ‘ + π’–π’–πŸ‘πŸ‘ + π’–π’–πŸ‘πŸ‘ + 𝒖𝒖 + 𝟏𝟏)

π’–π’–πŸπŸπŸ‘πŸ‘ + π’–π’–πŸ—πŸ— βˆ’ π’–π’–πŸ•πŸ• + π’–π’–πŸ–πŸ– βˆ’ π’–π’–πŸ‘πŸ‘ βˆ’ 𝟏𝟏

4. Beatrice writes down every expression that appears in this problem set, one after the other, linking them with β€œ+” signs between them. She is left with one very large expression on her page. Is that expression a polynomial expression? That is, is it algebraically equivalent to a polynomial?

Yes.

What if she wrote β€œβˆ’β€ signs between the expressions instead?

Yes.

What if she wrote β€œΓ—β€ signs between the expressions instead?

Yes.

Lesson 9: Multiplying Polynomials

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M1 Mid-Module Assessment Task

ALGEBRA I

8. a. Find the following products:

i. (π‘₯π‘₯ βˆ’ 1)(π‘₯π‘₯ + 1)

ii. (π‘₯π‘₯ βˆ’ 1)οΏ½π‘₯π‘₯2 + π‘₯π‘₯ + 1οΏ½ iii. (π‘₯π‘₯ βˆ’ 1)(π‘₯π‘₯3 + π‘₯π‘₯2 + π‘₯π‘₯ + 1) iv. (π‘₯π‘₯ βˆ’ 1)(π‘₯π‘₯4 + π‘₯π‘₯3 + π‘₯π‘₯2 + π‘₯π‘₯ + 1) v. (π‘₯π‘₯ βˆ’ 1)(π‘₯π‘₯𝑛𝑛 + π‘₯π‘₯π‘›π‘›βˆ’1 +β‹―+ π‘₯π‘₯3 + π‘₯π‘₯2 + π‘₯π‘₯ + 1)

b. Substitute π‘₯π‘₯ = 10 into each of the products from parts (i) through (iv) and your answers to show

how each of the products appears as a statement in arithmetic.

c. If we substituted π‘₯π‘₯ = 10 into the product (π‘₯π‘₯ βˆ’ 2)(π‘₯π‘₯7 + π‘₯π‘₯6 + π‘₯π‘₯5 + π‘₯π‘₯4 + π‘₯π‘₯3 + π‘₯π‘₯2 + π‘₯π‘₯ + 1) and computed the product, what number would result?

Module 1: Relationships Between Quantities and Reasoning with Equations and Their Graphs

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Page 18: Lesson 8: Adding and Subtracting Polynomials

M1 Mid-Module Assessment Task

ALGEBRA I

d. Multiply (π‘₯π‘₯ βˆ’ 2) and (π‘₯π‘₯7 + π‘₯π‘₯6 + π‘₯π‘₯5 + π‘₯π‘₯4 + π‘₯π‘₯3 + π‘₯π‘₯2 + π‘₯π‘₯ + 1), and express your answer in standard form. Substitute π‘₯π‘₯ = 10 into your answer, and see if you obtain the same result that you obtained in part (c).

e. Francois says (π‘₯π‘₯ βˆ’ 9)(π‘₯π‘₯7 + π‘₯π‘₯6 + π‘₯π‘₯5 + π‘₯π‘₯4 + π‘₯π‘₯3 + π‘₯π‘₯2 + π‘₯π‘₯ + 1) must equal π‘₯π‘₯7 + π‘₯π‘₯6 + π‘₯π‘₯5 + π‘₯π‘₯4 + π‘₯π‘₯3 +π‘₯π‘₯2 + π‘₯π‘₯ + 1 because when π‘₯π‘₯ = 10, multiplying by β€œπ‘₯π‘₯ βˆ’ 9” is the same as multiplying by 1.

i. Multiply (π‘₯π‘₯ βˆ’ 9)(π‘₯π‘₯7 + π‘₯π‘₯6 + π‘₯π‘₯5 + π‘₯π‘₯4 + π‘₯π‘₯3 + π‘₯π‘₯2 + π‘₯π‘₯ + 1). ii. Put π‘₯π‘₯ = 10 into your answer.

Is it the same as π‘₯π‘₯7 + π‘₯π‘₯6 + π‘₯π‘₯5 + π‘₯π‘₯4 + π‘₯π‘₯3 + π‘₯π‘₯2 + π‘₯π‘₯ + 1 with π‘₯π‘₯ = 10? iii. Was Francois right?

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Page 19: Lesson 8: Adding and Subtracting Polynomials

M1 Mid-Module Assessment Task

ALGEBRA I

8. a. Find the following products:

i. (π‘₯π‘₯ βˆ’ 1)(π‘₯π‘₯ + 1)

x2 + x βˆ’ xβˆ’ 1 x2-1

ii. (π‘₯π‘₯ βˆ’ 1)οΏ½π‘₯π‘₯2 + π‘₯π‘₯ + 1οΏ½ x3 + x2 + x βˆ’ x2 βˆ’ xβˆ’ 1 x3 βˆ’ 1

iii. (π‘₯π‘₯ βˆ’ 1)(π‘₯π‘₯3 + π‘₯π‘₯2 + π‘₯π‘₯ + 1) x4 + x3 + x2 + x βˆ’ x3 βˆ’ x2 βˆ’ x βˆ’1 x4 βˆ’ 1

iv. (π‘₯π‘₯ βˆ’ 1)(π‘₯π‘₯4 + π‘₯π‘₯3 + π‘₯π‘₯2 + π‘₯π‘₯ + 1) x5 + x4 + x3 + x2 + x βˆ’ x4 βˆ’ x3 βˆ’ x2 βˆ’ x βˆ’ 1 x5 βˆ’ 1

v. (π‘₯π‘₯ βˆ’ 1)(π‘₯π‘₯𝑛𝑛 + π‘₯π‘₯π‘›π‘›βˆ’1 +β‹―+ π‘₯π‘₯3 + π‘₯π‘₯2 + π‘₯π‘₯ + 1) xn+1 βˆ’ 1

b. Substitute π‘₯π‘₯ = 10 into each of the products from parts (i) through (iv) and your answers to show

how each of the products appears as a statement in arithmetic.

i. (10 βˆ’ 1) βˆ™ (10 + 1) = (100 βˆ’ 1) 9 βˆ™ (11) = 99

ii. (10 βˆ’ 1) βˆ™ (100 + 10 + 1) = (1000βˆ’ 1) 9 βˆ™ (111) = 999

iii. (10 βˆ’ 1) βˆ™ (1,000 + 100 + 10 + 1) = (10,000 βˆ’ 1) 9 βˆ™ (1,111) = 9,999

iv. (10 βˆ’ 1) βˆ™ (10,000 + 1,000 + 100 + 10 + 1) = (100,000 βˆ’ 1) 9 βˆ™ (11,111) = 99,999

c. If we substituted π‘₯π‘₯ = 10 into the product (π‘₯π‘₯ βˆ’ 2)(π‘₯π‘₯7 + π‘₯π‘₯6 + π‘₯π‘₯5 + π‘₯π‘₯4 + π‘₯π‘₯3 + π‘₯π‘₯2 + π‘₯π‘₯ + 1) and

computed the product, what number would result?

8 βˆ™ (11,111,111) = 88,888,888

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brittjm
Sticky Note
Answer key
Page 20: Lesson 8: Adding and Subtracting Polynomials

M1 Mid-Module Assessment Task

ALGEBRA I

d. Multiply (π‘₯π‘₯ βˆ’ 2) and (π‘₯π‘₯7 + π‘₯π‘₯6 + π‘₯π‘₯5 + π‘₯π‘₯4 + π‘₯π‘₯3 + π‘₯π‘₯2 + π‘₯π‘₯ + 1) and express your answer in standard form.

x8 + x7 + x6 + x5 + x4 + x3 + x2 + x βˆ’ 2x7 βˆ’ 2x6 βˆ’ 2x5 βˆ’ 2x4 βˆ’2x3 βˆ’2x2 βˆ’ 2xβˆ’ 2 x8 βˆ’ x7 βˆ’ x6 βˆ’ x5 βˆ’ x4 βˆ’ x3 βˆ’ x2 βˆ’ x βˆ’2

Substitute π‘₯π‘₯ = 10 into your answer and see if you obtain the same result as you obtained in part (c).

108 βˆ’107 βˆ’106 βˆ’105 βˆ’ 104 βˆ’ 103 βˆ’ 102 βˆ’ 10βˆ’ 2 = 88,888,888. Yes, I get the same answer.

e. Francois says (π‘₯π‘₯ βˆ’ 9)(π‘₯π‘₯7 + π‘₯π‘₯6 + π‘₯π‘₯5 + π‘₯π‘₯4 + π‘₯π‘₯3 + π‘₯π‘₯2 + π‘₯π‘₯ + 1) must equal π‘₯π‘₯7 + π‘₯π‘₯6 + π‘₯π‘₯5 + π‘₯π‘₯4 + π‘₯π‘₯3 +π‘₯π‘₯2 + π‘₯π‘₯ + 1 because when π‘₯π‘₯ = 10, multiplying by "π‘₯π‘₯ βˆ’ 9" is the same as multiplying by 1.

i. Multiply (π‘₯π‘₯ βˆ’ 9)(π‘₯π‘₯7 + π‘₯π‘₯6 + π‘₯π‘₯5 + π‘₯π‘₯4 + π‘₯π‘₯3 + π‘₯π‘₯2 + π‘₯π‘₯ + 1).

x8 βˆ’ 8x7 βˆ’ 8x6 βˆ’ 8x5 βˆ’ 8x4 βˆ’8x3 βˆ’8x2 βˆ’ 8xβˆ’ 9

ii. Put π‘₯π‘₯ = 10 into your answer.

100,000,000βˆ’ 80,000,000βˆ’ 8,000,000βˆ’ 800,000βˆ’ 80,000 βˆ’8,000 βˆ’ 800βˆ’ 80βˆ’ 9

100,000,000βˆ’ 88,888,889 = 11,111,111

Is it the same as π‘₯π‘₯7 + π‘₯π‘₯6 + π‘₯π‘₯5 + π‘₯π‘₯4 + π‘₯π‘₯3 + π‘₯π‘₯2 + π‘₯π‘₯ + 1 with π‘₯π‘₯ = 10?

Yes.

iii. Was Francois right?

No, just because it is true when x is 10, doesn’t make it true for all real x. The two

expressions are not algebraically equivalent.

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