Lesson Plan

Synthetic Division and the Factor Theorem


This lesson is going to build on the concept of polynomials and their roots. Students will:

  • illustrate the relationship between roots and factors. [IS.3 - Struggling Learners]

  • use synthetic division to determine if a certain value is a root of a polynomial. [IS.4 - Struggling Learners]

Essential Questions

How are relationships represented mathematically?
How can expressions, equations, and inequalities be used to quantify, solve, model, and/or analyze mathematical situations?
How can mathematics support effective communication?
How can patterns be used to describe relationships in mathematical situations?
How is mathematics used to quantify, compare, represent, and model numbers?
What does it mean to estimate or analyze numerical quantities?
What makes a tool and/or strategy appropriate for a given task?
  • How can we determine if a real-world situation should be represented as a quadratic, polynomial, or exponential function?

  • How do you explain the benefits of multiple methods of representing polynomial functions (tables, graphs, equations, and contextual situations)?


  • Factor: One of two or more expressions that are multiplied. [IS.1 - Struggling Learners]

  • Fundamental theorem of algebra: Every polynomial equation of degree n ≥ 1, with complex coefficients, has at least one root which is a complex number (real or imaginary).

  • Root: The solution to a given equation.

  • Synthetic division: A short way of dividing a polynomial by a binomial.

  • Remainder: In division, the difference of the dividend minus the product of the quotient times the divisor.

  • Zero of a function: The value of the argument for which the function is equal to zero; also the Root of the function and the Solution to the equation. [IS.2 - Struggling Learners]


90–120 minutes [IS.5 - All Students]

Prerequisite Skills

Prerequisite Skills haven't been entered into the lesson plan.


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Formative Assessment

  • View
    • Student results from the Extension activity will reflect how well they are able to generalize the principles of synthetic division. Using literal terms can cause students to be distracted by the need for discipline in notation. For students who encounter difficulty in the activity, suggest substituting trial values with small integers to obtain binomial results. [IS.8 - Struggling Learners]

    • The Lesson 2 Exit Ticket activity (M-A2-3-2_Lesson 2 Exit Ticket.doc and M-A2-3-2_Lesson 2 Exit Ticket KEY.doc) requires students to match the unique roots of the polynomial with the corresponding expression.

Suggested Instructional Supports

  • View
    Active Engagement, Modeling, Explicit Instruction

    This lesson introduces students to synthetic division and models some techniques for testing whether or not a given term is a factor of a polynomial expression. The lesson also helps students to form their own representation of the relationship between the roots and the factors of polynomial expressions.

    Calculating the quotient with a multi-digit divisor is an appropriate introduction for understanding the techniques of synthetic division. The two processes are related in their structure and outcomes, and long division with a familiar algorithm gives students confidence in trying the new method.

    Divide, multiply, subtract, check, and bring down in the prescribed sequence outlines the operational steps for synthetic division and makes it possible for students to perform the operations with confidence.


    In Activity 3 (pair activity), students have to create their own polynomials by beginning with numbers they select as roots. Recognizing the roots in factored form has to involve thinking backwards from procedures they have previously practiced. Students must also evaluate the work of another student through their own understanding of the procedure.


    The Lesson 2 Exit Ticket helps students evaluate their individual understanding of synthetic division. Students must represent their understanding of the process by matching the unique roots to each polynomial.

    Group and partner work is used throughout so that students can help each other. Emphasis should be placed on communicating mathematical ideas with the specific vocabulary words appropriate to the concepts. Partnering activities require students to represent their individual understanding of the concept as well as evaluate the representations of another student. Student sharing of ideas and insights are helpful ways to increase learning without your intervention.

    Logical/Sequential/Visual Learners: It may be helpful for some students to see several written versions of the individual steps in synthetic division. Speak the name of the terms and operations appropriately while writing or pointing to them in each use.


    This lesson opens with a simple activity that grabs students’ attention because it illustrates the concept of shortcuts and easier processes, especially in math. This leads nicely into the topic of long and synthetic division. Concepts and vocabulary are explained and then students practice their skills. The class comes back together for more instruction and practices their skills in pair work. This lesson has a mixture of whole class, individual, and group work; it is designed so that each activity flows into the next.


    IS.1 - Struggling Learners

    Consider the following steps with regard to vocabulary for struggling learners:

    1. Use of a graphic organizer (e.g., Frayer Model, Verbal Visual Word Association, Concept Circles).
    2. Introduce new vocabulary using student friendly definitions and examples and non-examples.
    3. Review words with students.
    Provide opportunities for students to apply the new/reviewed terms. 
    IS.2 - Struggling Learners
    This use of the word argument may cause some problems for struggling learners.  Consider using a more student friendly definition and provide written examples.  
    IS.3 - Struggling Learners
    Consider providing struggling learners with examples of this.  
    IS.4 - Struggling Learners
    Consider reviewing and modeling this for struggling learners as well as providing written examples.  
    IS.5 - All Students
    Consider pre-teaching the concepts critical to this lesson.  Use formative assessments throughout the lesson to determine students level of understanding.  Use follow-up reinforcement if necessary.  
    IS.6 - Struggling Learners
    Consider having  the written step and the example for that step written under the verbiage for struggling learners.  It may be difficult for them to follow by having the verbiage for all the steps written above and the examples of each step written separately.  
    IS.7 - Struggling Learners
    See above comment.  
    IS.8 - Struggling Learners
    Consider using multiple forms of formative assessment for struggling learners.  

Instructional Procedures

  • View

    After this lesson, students will know what the factor theorem is and how it relates roots to factors. They will understand how synthetic division shows when a value is a root of a polynomial and that synthetic division produces the same results as long division but is much easier and quicker. This lesson builds on the basics of polynomials and introduces characteristics of a polynomial’s graph. Students need to understand where roots come from and will use their knowledge of roots in the application of polynomial functions in real-world situations. Students will be able to use synthetic division to determine if a value is a root. They will be able to write polynomial equations knowing the roots and write polynomials in factored form.

    Have students look at the graph of a parabola (for example: y = x2 + 2x − 3). Then have students answer the following questions:

    • What is the degree and how many x-intercepts does the graph have? (Refer to the fundamental theorem of algebra)”

    • What are the zeros or x-intercepts?”

    • What is the equation of the parabola in factored form?”

    • What connection is there between the factors and zeros?”

    Since x2 + 2x − 3 = (x + 3)(x − 1), that means l2-01fraction.PNGx − 1. We use division of polynomials to help determine the roots of the polynomials. How do we divide polynomials? We can use what we know from long division, or we can take the shortcut and use synthetic division.”


    Long Division:


    Step 1: x2 divided by x = x (line up like terms)  (x² / x) = x

    Step 2: Multiply x by x + 3 and write the product underneath the dividend.

    Step 3: Subtract the product by distributing the negative sign across all terms in the product.

    Step 4: − x divided by x = − 1 or (−x / x) =  −1

    Step 5: Multiply − 1 by x + 3 and write the product underneath; then subtract.


    [IS.6 - Struggling Learners]

    Step 5
    Second Example of Long Division:

    This process is like long division. We are going to use an easier process called synthetic division for dividing polynomials.”

    There is one requirement in order to use synthetic division: The divisor has to be of the form xa or x + a. Using the same example from above, divide x2 + 2x − 3 by x + 3.”

    The following notes should be displayed on the board for students to copy.

    Step 1: Write the coefficients of the dividend (the numbers in front of the variables and the constant: 1, 2, − 3) and draw an “L” around the numbers as well as a dashed line in front of the constant (optional).

    Step 2: Since the divisor is x + 3, because x + 3 = 0 and x = –3, we are going to use − 3 for synthetic division. Place − 3 on the left side of the 1 on the other side of the line.



    Step 3: Bring down the 1. (Always bring down the first coefficient.)


    Step 4: Multiply the 1 by the divisor − 3 and place the product under the 2.


    Step 5: Add 2 + −3 and write the sum under the line.

    Step 6: Multiply the − 1 by the divisor − 3 and place the product under the − 3.


    Step 7: Add − 3 + 3 and write the sum under the line.

    Step 8: Write out the quotient. The number on the right of the dashed line is the remainder. The first number to the left of the dashed line is the new constant, the next number is the coefficient in front of x, and if there were one more number, it would be the coefficient in front of x2 and so on. This process can be used with third, fourth, and larger degree polynomials as long as the divisor is of the form x ± a.


    [IS.7 - Struggling Learners]

    Third example for synthetic division:

    Note: In Step 2, explain that in synthetic division you use the opposite of the sign in the divisor. Remember that the sum of the constant term and its additive inverse must be zero. Our divisor was x + 3, so in synthetic division it needs to be − 3. If we had a divisor of x − 4, we would use + 4. Stop and see if anyone has any questions at this step.

    Activity 1: Group Work

    If the classroom has whiteboards on all four walls, write the following division problems on the boards. If the room doesn’t have whiteboards all the way around, place the problems on poster paper and hang them around the room.

    Long Division: 1.(x3 + x2 − 5x − 2) ÷ (x − 2)

    2. (x2 + 2x + 1) ÷ (x + 1)

    3. (x3 − 3x2 − 11x + 5) ÷ (x2 + 2x − 1)

    4. (x4 + 4x3 + 6x2 + 4x + 1) ÷ (x + 1)

    Answers: 1. x2 + 3x + 1

    2. x + 1

    3. ( – 5)

    4. x3 + 3x2 + 3x + 1

    Synthetic Division:

    5. (9x3 − 6x2x +30) ÷ (x − 3)

    6. (x2 − 3x − 28) ÷ (x + 4)

    7. (x3x2 − 2x) ÷ (x + 1)

    8. (x3 + 7x2 − 38x + 40) ÷ (x − 2)

    9. (x4 + 3x3 − 25x2 + 33x + 72) ÷ (x + 6)

    10. (x2 + x − 72) ÷ (x − 8)

    Answers: 5. (9x2 + 21x + 62) + (216 / x – 3)

    6. (x – 7)

    7. (x3 – 2x)

    8. (x2 + 9x – 20)

    9. (x3 – 3x2 – 7x + 75) – (378 / x + 6)

    10. (x + 9)


    Split the class into groups of three to four and have each group go to a problem on the board. Have students work on the problem in front of them for one to two minutes. If they finish the problem in that time, they should raise their hands and have you check their work. If they are right, they should erase their work so the next group doesn’t see it. If they do not finish the problem, that’s okay. After time is up, have the class rotate clockwise. The groups then have to look at the next problem and see if the group in front of them was on the right track to solve the problem. The current group finishes the problem and gets the work checked. Repeat this process for 10 to 15 minutes, making sure each group has seen at least one long-division problem and about four to five synthetic-division problems.

    Bring the class back together. Ask, “Which process did you like better and why?”

    Most likely they will ask why synthetic division works. This is a good time to introduce the Factor Theorem. “Since polynomials can be used to model real-world applications, it is important to determine the x-intercepts for information about the real-world situation.”

    Put the following on the board for students to copy in their notes.

    Factor theorem: If a is used to synthetically divide a polynomial and it produces a remainder of zero, then not only is x = a a root of the polynomial, but xa is a factor of the polynomial.

    In the previous example, −3 was used in synthetic division with the polynomial x2 + 2x − 3 and it produced a remainder of 0; therefore, x − (−3) or x + 3 is a factor of the polynomial.

    Synthetic division is a step-by-step process where each step gives information that is useful for the next step. In that way, it builds upon itself.”

    Example: Determine which of the following are roots of the polynomial x3 + 5x2x − 5: 1, − 1, 2, and − 5.


    Example: (x3 – 5x– 4) ÷ (x – 3)
    3     1     0     –5     –4
                  3      9     12
           1     3     4     8

    Answer: x2 + 3x + 4 + ( 8 / x – 3 )


    Activity 2: Pairs

    Place the following table on the board and discuss with the class the middle column.


    Number of Roots the Polynomial Should Have

    Possible Roots

    x3 + 3x2 − 6x − 8


    2, 1, − 4, − 1

    x4 − 34x2 + 225


    −5, 5, 3, − 3, 2

    x2 + 3x − 28


    4, − 7, 6

    x2 + 2x


    0, − 2, 3

    With your partner, you are going to determine which numbers are roots and which numbers are not roots. You should use synthetic division as in the previous example. In the second polynomial, you will need to use a coefficient of 0 for the x3 and x terms.”




    x3 + 6x2 + 5x G


    x2 − 7x + 6 E


    x4 − 7x3 − 7x2 + 43x + 42 F


    x2 − 10x + 16 B


    x4 + 4x3 − 53x2 − 168x C


    x5 + x4 − 51x3 + 31x2 + 538x − 840 H


    x2 + 10x + 24 A


    x3 − 15x2 + 74x − 120 D




    −4, −6


    2, 8


    −3, −8, 0, 7


    4, 5, 6


    1, 6


    −2, 7, −1, 3


    −5, −1, 0


    3, −7, 2, −4, 5











    Walk around and check students’ work. Have students who are catching on quickly write the polynomials in factored form. From the previous example, if they determine that −1, 1, and −5 are roots of the polynomial x3 + 5x2x − 5, then they can write the roots as factors: x3 + 5x2x − 5 = (x + 1)(x − 1)(x + 5). Choose a couple of groups to write their work on the board. This is a good way for students to check their work and ask questions.

    Activity 3:

    Individually come up with your own polynomials by following these steps:

    1. Think of three to five numbers to be the roots of a polynomial.

    2. Write those roots in factored form.

    3. Multiply all the factors together to create a polynomial in standard form.”

    Next, have students pass their polynomial to the student behind them along with two of the roots and a number that is not a root. Students then have to determine the roots and the nonroots. When everyone is finished, students pass their work back to the person in front of them to check for errors.

    Another example: (x + 7)(x – 6)(x – 11)

    x3–10x2–53x + 462 possible roots, –2, –1, 1, 2


    Use the Lesson 2 Exit Ticket (see M-A2-3-2_Lesson 2 Exit Ticket and M-A2-3-2_Lesson 2 Exit Ticket KEY in the Resources folder) to evaluate students’ understanding.

    Use synthetic division to match the polynomials on the left with their roots on the right.


    • If x = a is a root of each polynomial, then find the other binomial factor of the following trinomials.

          1. x2ax + 3x – 3a          (x + 3)

          2. x2ax – 5x + 5a                (x – 5)

          3. x2ax + bxab                (x + b)

          4. x2axbcx + abc          (x bc)

          5. 6x3x2 – 13x + 3           (2x + 3)

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