# Lesson 15

The Remainder Theorem

## 15.1: Notice and Wonder: Division Leftovers (5 minutes)

### Warm-up

The purpose of this warm-up is to focus student attention on the link between the divisor, dividend, quotient, and remainder and how those 4 terms can be used to write a multiplication equation whether they involve numbers or polynomials. Of particular importance is that not all division works out “evenly” and the meaning of the remainder, an idea touched on briefly in the previous lesson. In the following activities, students will translate this thinking to the work of dividing polynomials and the meaning of the remainder.

### Launch

Display the long division and equations for all to see. Ask students to think of at least one thing they notice and at least one thing they wonder. Give students 1 minute of quiet think time, and then 1 minute to discuss the things they notice and wonder with their partner, followed by a whole-class discussion.

### Student Facing

What do you notice? What do you wonder?

### Student Response

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### Activity Synthesis

Ask students to share the things they noticed and wondered. Record and display their responses for all to see. If possible, record the relevant reasoning on or near the image. After all responses have been recorded without commentary or editing, ask students, “Is there anything on this list that you are wondering about now?” Encourage students to respectfully disagree, ask for clarification, or point out contradicting information.

If the difference between 0 and non-0 remainders does not come up during the conversation, ask students to discuss this idea.

## 15.2: The Unknown Coefficient (10 minutes)

### Activity

While the link from seeing the factor \((x-a)\) to stating that \(x=a\) is a zero of a function has been made in several previous lessons, this activity asks students to use that knowledge in a novel way to identify an unknown coefficient in an expression for a polynomial function. Additionally, as a result of this activity, students should have a clear understanding of why a polynomial with a known factor of \((x-a)\) is equal to zero when \(x=a\) using the relationships between division and multiplication equations. Monitor for students using different ways to identify the value of \(u\), such as:

- graphing \(y=f(x)\) for different values of \(u\) to find a graph with \((2,0)\) as a horizontal intercept
- using long division and the knowledge that the remainder after dividing by \((x-2)\) is 0
- substituting 2 for \(x\) and solving for \(u\) when the expression is equal to 0

### Launch

*Representing, Conversing: MLR7 Compare and Connect.*Use this routine to prepare students for the whole-class discussion. At the appropriate time, ask students to prepare a visual display that shows their mathematical thinking and reasoning for the given question. Invite students to quietly circulate and read at least 2 other posters or visual displays in the room. Give students quiet think time to consider what is the same and what is different about the displays. Next, ask students to find a partner to discuss what they noticed. Listen for and amplify observations involving the strategy of substituting 2 for \(x\). This will help students connect their understanding of linear factors, zeros of a function, and function notation as they discuss different strategies.

*Design Principle(s): Cultivate conversation*

*Representation: Internalize Comprehension.*Differentiate the degree of difficulty by beginning with an example with more accessible values. Highlight connections between representations by starting with a simpler problem, such as \(x^2 + ux + 10\), with a known factor of \(x+ 5\) (\(u=7\)).

*Supports accessibility for: Conceptual processing*

### Student Facing

Consider the polynomial function \(f(x)=x^4 - ux^3 + 24x^2 - 32x + 16\) where \(u\) is an unknown real number. If \(x-2\) is a factor, what is the value of \(u\)? Explain how you know.

### Student Response

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### Student Facing

#### Are you ready for more?

Here are some diagrams that show the same third-degree polynomial, \(P(x) = 2x^3+5x^2+x+10\), divided by a linear factor and by a quadratic factor.

\(\dfrac{P(x)}{x+3}\)

\(2x^2\) | \(\text-x\) | 4 | |

\(x\) | \(2x^3\) | \(\text-x^2\) | \(4x\) |

3 | \(6x^2\) | \(\text-3x\) | 12 |

\(\dfrac{P(x)}{x^2-x}\)

\(2x\) | 7 | |

\(x^2\) | \(2x^3\) | \(7x^2\) |

\(\text-x\) | \(\text-2x^2\) | \(\text-7x\) |

- What is the remainder of each of these divisions?
- For each division, how does the degree of the remainder compare to the degree of the divisor?
- Could the remainder ever have the same degree as the divisor, or a higher degree? Give an example to show that this is possible, or explain why it is not possible.

### Student Response

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### Activity Synthesis

Select previously identified students to share how they calculated the value of \(u\) in the order previously listed. Record student responses for all to see.

The third strategy, substituting 2 for \(x\), is particularly important to reach because students using the first strategy could just be remembering the pattern that a polynomial with \((x-a)\) as a factor intercepts the horizontal axis at \(a\) and students may not really be thinking about the algebraic meaning of horizontal intercepts. Students using the second strategy, long division, could just be remembering that if you divide by a factor, you get a remainder of 0. Only students using the third strategy are explicitly setting \(f(2)\) equal to 0 and connecting their understanding of linear factors of a polynomial, the zeros of a function, and function notation.

In order to encourage students to think further about this assumption, ask, “How do you know that \(f(2)=0\) since \((x-2)\) is a factor of the polynomial?” Encourage students to try and answer the question without referencing the graph of the function and instead consider how to prove it using what they know about the relationship between division and remainders and writing multiplication equations. After some quiet think time, invite students to share their ideas. If no student suggests it, ask them how they could write out the relationship between \(f(x)\) and \((x-2)\) like the relationship between \(330\) and the various integers in the warm-up. Students should understand that since \(f(x)= (x-2)(C) + 0\), where \(C\) is a cubic function, we then also know that \(f(2)=0\), because \(f(2)=(2-2)(C)+0=0\).

## 15.3: A Study of Remainders (20 minutes)

### Activity

The purpose of this task is to establish the Remainder Theorem and from it, conclude that for a polynomial \(p(x)\), if \(p(a)=0\), then \((x-a)\) is a factor of \(p(x)\). In this activity, students are generalizing, so graphing technology is not an appropriate tool. Instead, encourage students to think of other strategies they can use to answer the questions.

### Launch

Arrange students in groups of 2–3. After 5 minutes of work time, pause the class and ask 2–3 groups to share their reasoning about whether \((x-2)\) is a factor of each polynomial. In particular, highlight why testing the value of the function at \(x=2\) works, building off the conclusion of the previous activity. Encourage students to try this activity without using graphing technology, but offer it as a support for those who need it.

*Engagement: Internalize Self Regulation.*Chunk this task into more manageable parts to differentiate the degree of difficulty or complexity. Invite students to choose and respond to 3 of the 6 polynomials.

*Supports accessibility for: Organization; Attention*

### Student Facing

- Which of these polynomials could have \((x-2)\) as a factor?
- \(A(x) = 6x^2 - 7x - 5\)
- \(B(x) = 3 x^2 + 15 x - 42\)
- \(C(x) = 2x^3 + 13x^2 + 16x + 5\)
- \(D(x) = 3x^3 - 2x^2 - 15x + 14\)
- \(E(x) = 8x^4 - 41x^3 - 18x^2 + 101x + 70\)
- \(F(x) = x^4 + 5 x^3 - 27 x^2 - 101 x - 70\)

- Select one of the polynomials that you said doesn’t have \((x-2)\) as a factor.
- Explain how you know \((x-2)\) is not a factor.
- If you have not already done so, divide the polynomial by \((x-2)\). What is the remainder?

- List the remainders for each of the polynomials when divided by \((x-2)\). How do these values compare to the value of the functions at \(x=2\)?

### Student Response

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### Anticipated Misconceptions

Students who do not yet feel fluent with polynomial division may find it helpful to double-check their conclusions about which polynomials have \((x-2)\) as a factor by graphing them to see which ones have a zero at \(x=2\).

### Activity Synthesis

The focus of this discussion is on the results of the last question, in which students found that the remainders when dividing the list of polynomials by \((x-2)\) are all the same as the value of the polynomials at \(x=2\). Ask students, “Why does it seem like the remainder when \(p(x)\) is divided by \((x-a)\) is equal to \(p(a)\)?” Give students 2–3 minutes work time. If needed, directly encourage students to use the relationship between division and multiplication that they have seen in the activities leading up to this one. Invite students to explain what they think is happening.

It is important for students to understand that since all division problems can be rewritten as multiplication problems, we can think of dividing the polynomial \(p(x)\) by the linear factor \((x-a)\) as \(p(x) = (x-a)q(x)+r\), where \(r\) is the remainder and \(q(x)\) is a polynomial. When \(x=a\), \(p(a)=(a-a)q(x)+r=r\), so the remainder after division by \((x-a)\) is \(p(a)\). Tell students that this is called the R*emainder Theorem*.

This theorem allows us to state that if we have a polynomial \(p(x)\) with a known zero at \(x=a\), then \((x-a)\) is a factor of \(p(x)\) since if we divide \(p(x)\) by \((x-a)\), then we have \(p(x) = (x-a)q(x)+r\), where \(r\) is the remainder and \(q(x)\) is a polynomial. Because we already know \(a\) is a zero of the function, we know that \(p(a)=0\). This means we also know that the remainder is zero, since \(p(a) = (a-a)q(x)+r=r=0\). So \(p(x) = (x-a)q(x)\) if \(x=a\) is a zero of the polynomial.

*Writing, Listening, Conversing: MLR1 Stronger and Clearer Each Time.*Use this routine to help students improve their verbal and written responses to the question “Why does it appear that the remainder when \(p(x)\) is divided by \((x-a)\) is equal to \(p(a)\)?” Give students time to meet with 2–3 partners to share and get feedback on their responses. The listener should press for detail and precision in language by asking, “Can you give an example?”, “How do you know?”, or “Is there a different way you can explain that?” Invite students to go back and revise or refine their written explanation based on the feedback from peers. This will help students analyze remainders when dividing polynomials through conversations with multiple partners.

*Design Principle(s): Optimize output (for explanation); Cultivate conversation*

## Lesson Synthesis

### Lesson Synthesis

The purpose of this discussion is to give students a chance to demonstrate what they have learned so far this unit. Display for all to see “The answer is \(f(x)=(2x-1)(x+2)(x-8)\). What was the question?” Here are some possible responses.

- Write the equation for a 3rd degree polynomial.
- Write the equation for a polynomial where \(f(\text-2)=f(0.5)=f(8)=0\) and the vertical intercept is \((0, 16)\).
- Rewrite this equation in factored form: \(f(x)=2x^3 - 13x^2 - 26x + 16\).
- Use division to find the polynomial that makes the following equation true: \(2x^4 - 11x^3 - 39x^2 - 10x + 16 =(x+1)f(x)\).
- What value for \(a\) makes \(f(0.5)=0\) true if \(f(x)=(ax-1)(x+2)(x-8)\)?

Invite students to share their questions, recording them for all to see.

## 15.4: Cool-down - Using Remainder Knowledge (5 minutes)

### Cool-Down

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## Student Lesson Summary

### Student Facing

When we use long division to divide 1573 by 12, we get a remainder of 1, so \(1573 = 12(131) + 1\). When we divide by 11 instead, we get a remainder of 0, so \(1573 = 11(143)\). A remainder of 0 means that 11 is a factor of 1573. The same thing happens with polynomials. While \((x^3 + 5x^2 + 7x + 3) \div (x+2)\) results in a remainder that is not 0, if we divide \((x+1)\) into \(x^3 + 5x^2 + 7x + 3\), we do get a remainder of 0:

\(\displaystyle \require{enclose} \begin{array}{r} x^2+4x+3 \\ x+1 \enclose{longdiv}{x^3+5x^2+7x+3} \\ \underline{\text-x^3-x^2} \phantom{+7x+333} \\ 4x^2+7x \phantom{+33}\\ \underline{\text-4x^2-4x} \phantom{+33} \\ 3x+3 \\ \underline{\text-3x-3} \\ 0 \end{array}\)

So \((x+1)\) is a factor of \(x^3 + 5x^2 + 7x + 3\).

Earlier we learned that if \((x-a)\) is a factor of a polynomial \(p(x)\), then \(p(a)=0\), meaning \(a\) is a zero of the function. It turns out that the converse is also true: if \(a\) is a zero, then \((x-a)\) is a factor.

To see that this is true, let’s think about what we know if we have a polynomial \(p(x)\) with a known zero at \(x=a\). If we divide \(p(x)\) by the linear factor \((x-a)\), then \(p(x) = (x-a)q(x)+r\), where \(r\) is the remainder and \(q(x)\) is a polynomial. Because \(a\) is a zero of the function, we know that \(p(a)=0\). This means we also know that the remainder is zero:

\(\displaystyle p(a) = (a-a)q(x)+r \\ p(a)=r\\ 0=r\)

Which means that \(p(x) = (x-a)q(x)\). So, if \(a\) is a zero of a polynomial, then \((x-a)\) must be a factor of \(p(x)\). Now we know that if we start with a linear factor of a polynomial, then we know one of the zeros of the polynomials, and if we start with a zero of a polynomial, then we know one of the linear factors.

Lastly, even if \(a\) is not a zero of \(p\), we can figure out what the remainder will be if we divide \(p(x)\) by \((x-a)\), without having to do any division. If \(p(x) = (x-a)q(x)+r\), then \(p(a)=(a-a)q(x)+r\), so \(p(a)=r\). So the remainder after division by \((x-a)\) is \(p(a)\). This is the Remainder Theorem.