The purpose of this lesson is for students to understand the Remainder Theorem. A focus of the lesson is that for any polynomial \(p(x)\) and linear expression \((x-a)\), we can state that \(p(x) = (x-a)q(x) + r\), where \(r\) is the remainder and \(q(x)\) is a polynomial. To get to this endpoint, the start of the lesson asks students to consider the meaning of the remainder when dividing with integers, explicitly making the connection between dividend, divisor, quotient, and remainder written using long division and written as a multiplication equation.
In the following activity, students identify an unknown coefficient in a polynomial that has a known factor. Students use a consequence of the Remainder Theorem studied previously: if a polynomial \(p(x)\) has a factor of \((x-a)\), then \(p(a)=0\). Using new connections made during their work with division and writing multiplication equations, students now consider this fact from the perspective of what must be true about \(p(x)=(x-a)q(x)+r\) when we know \(r=0\). This activity is purposely unscaffolded to encourage students to make sense of the problem and choose a solution path (MP1).
In the last activity, students complete repeated calculations to help them make the connection that the value of \(p(a)\) is equal to the value of the remainder when \(p(x)\) is divided by \((x-a)\) (MP8). Once the Remainder Theorem is established, it can then be stated that, for a polynomial \(p(x)\), \(p(a)=0\) means \((x-a)\) must be a factor. In previous lessons, students used zeros to predict factors, and now they will know that zeros always correspond to factors in this way.
- Comprehend that for a polynomial $p(x)$, $(x-a)$ is a factor if $p(a)=0$ and, conversely, that $p(a)=0$ if $(x-a)$ is a factor.
- Comprehend that for a polynomial $p(x)$ and a number $a$, the remainder on division by $(x - a)$ is $p(a)$.
- Let’s learn about the Remainder Theorem.
- I understand the remainder theorem and why it's true.
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