Lesson 15

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.

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.

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.

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

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 \(f(2)=(2-2)(C)+0=0\).

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.

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.

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 \(x=2\).

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 Remainder 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.