Experience First

We are taking a new approach to dividing polynomials with this lesson.  Instead of getting into the step-by-step procedure of long division and synthetic division, we want the students to really understand 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 model to demonstrate how to multiply two polynomials. Students are familiar with this approach from Algebra 2. Students should notice be able to identify like terms and combine them to answer the question.  They will use this same method, but backwards, to divide in #3 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 3x^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, you will get 6 in the bottom right corner where the constant usually goes, you have to ask yourself, “How can I adjust my answer so that (x+3)(x^2+x+2) = (x^3+4x^2+5x+2)?" The 6 needs to become a 2!”  From there, you will subtract the 4. Since the remainder is -4, the quotient is x^2+x+2 -4/(x+3).

For some easier context, you can explain to the students that if I divide 32 by 5 with this method, 5 goes into 32 six whole times, but there is some left over. 32=5(6)+2. This means that 32/5 = 6 + 2/5, or 6.4. Students should feel comfortable writing polynomial products and quotients in this way.

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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 (which 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 (x-3) and the values of f(x) when x = -1 and x = 3.  This is the Remainder Theorem, which states that if (x-k) is a factor of f(x), then f(x)/(x-k) has a remainder of 0.  It also goes further to say that the remainder when dividing a polynomial f(x) by any (x-k) is equal to f(k).