Factor and Remainder Theorem (Lesson 2.3 Day 1)
Unit 2  Day 6
Unit 2
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Day 2
Day 3
Day 4
Day 5
Day 6
Day 7
Day 8
Day 9
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Day 11
Day 12
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Learning Objectives

Explain why (xk) is a factor of the polynomial is x=k is a zero of the polynomial

Interpret the remainder of a polynomial divided by (xk)

Given one factor of a polynomial function, use division to fin the remaining factors
Quick Lesson Plan
Experience First
We are taking a new approach to dividing polynomials with this lesson. Instead of getting into the stepbystep procedure of long division and synthetic division, we want the students to really understanding what they’re doing when they are dividing one polynomial by another. One way to think of division is “what would we multiply this by to produce the solution?” which is how we want the students to approach this topic.
We start with the area box method to demonstrate how to multiply two polynomials using the box. The students should notice how to find the like terms from the box and combine them to answer the question. They will this same method, but backwards, to divide in #2 and #5.
Many will struggle at first, but remind them that once you know one of the numbers in the top, you can multiply it by the other side to get more values on the inside. Then they have to think “x^2 has to combine with something to get 4x^2” and from there they can fill in the 5x^2. The whole process is like a puzzle where you’re working backwards and forwards until you have the final answer.
Number 5 is more challenging because it does not divide evenly. For this problem, once you get 12 in the box where the constant usually goes, you have to ask yourself, “How can I adjust my answer so that (x3)(x^2x4) = (x^34x^2x+4). The 12 needs to become a 4!” From there, you will subtract the 8. But remember—it should be 8/(x3) so that it’s mathematically sound.
For some easier context, you can explain to the students that if I divide 32 by 5 with this method, 5 goes into 32 a total of 6 times, but I have to add in the extra 2/5 to make it completely correct. 32/5 = 6 + 2/5, or 6.4.
Formalize Later
We are trying to move away from “follow these steps to get your answer” math. Emphasize trying to reason your way to the solution instead of teaching a tedious method like dividing polynomials using long division or a “fake” method like synthetic (literally means fake) division. The margin notes are minimal for the dividing portion for this reason. However, it’s very important to emphasize that when there is a remainder, you must divide it by the divisor when writing the final answer.
In #7 and #8, we hope they see the connection between the remainders for (x+1) and (x3) and the values of f(x) when x = 1 and x = 3. This is the Remainder Theorem, which states that if (xk) is a factor of f(x), then f(x)/(xk) has a remainder of 0. It also goes further to say that the remainder when dividing a polynomial f(x) by any (xk) is equal to f(k).