Based on observations students have already made in their previous work with cubics, it's time to formalize the Factor Theorem.
I will begin with Two Preludes to the Factor Theorem. First, I'll write the polynomial (in expanded form) on the board and ask if anyone can identify its roots for me. After about 5 seconds of silence, I'll write the factored form and ask, "How about now?" I expect many students will be able to explain that each factor corresponds to a root, and they should be able to name the root. Second, I'll sketch a polynomial with several integer roots and ask what we can say about its equation. I expect the students to be ready at this point to begin naming the binomial factors implied by the roots. This little introduction sets the stage for the formal statement of the Factor Theorem and its converse.
I've included some notes that guide my explanation, but the idea is to build on the multiplication property of zero to show that if (x - b) is a factor, then f(b) = 0. We aren't in a position to prove the converse, and I'm not sure it's worth the time, so we'll assume that it's also true that if f(b) = 0, then (x - b) is a factor of the polynomial. (MP 7)
Of course this all begs the question of how this theorem helps us. Practically, it says that we can use the factored form to predict the x-intercepts of the graph, and that we can use the x-intercepts to find the factored form of the equation. In other words, although factoring a polynomial is difficult, the graph can make it easier.
But making the graph is difficult, too, because we have to evaluate the function at so many points!
True. But if we could find a quicker way to evaluate the function, then graphing would be easier, and so would factoring...
Now, since we live in the 21st Century, we can just use a computer (or a phone, for that matter) to solve all of these graphing, factoring, & evaluating problems for us. But let's put ourselves in the shoes of the mathematicians who first struggled with all of this. That takes us back to Europe in the late 1700's and early 1800's (can anyone remember what was going on in America at that time?). By that time, mathematicians had found general solutions to the quadratic and cubic functions, but not the quartic. In their search, they had to evaluate many different equations for many different values. Clearly, any shortcut would be welcome!
I now introduce (see my notes) the "nested form" of a quartic equation and explain the advantage of using this form to evaluate the function. The steps are simple and repetitive.
Paolo Ruffini (ca. 1809) noticed this, too, and turned it into the method we call "Synthetic Substitution". Now it's important to explain synthetic substitution with constant reference to the steps taken in the nested form of the equation. If you think of synthetic substitution as a machine, then the nested form of the equation reveals exactly how the machine works, and why. (MP 7 & 8)
After the class example is completed and the synthetic substitution procedure has been explained and modeled, it's time for the students to get some practice. Hand out Synthetic Substitution Practice.
When they begin, I'll ask the students to take 2 minutes INDIVIDUALLY to complete first synthetic substitution problem. There will probably be a number of false starts, so I'll follow up with about 2 minutes for questions, but I'll try to have other students in the class provide all the answers. Then I'll give them another 2 minutes to work INDIVIDUALLY, and during this time I'll make a point of checking in with any students who are still unsure of the procedure.
Finally, once I'm sure everyone has understood how to do the synthetic substitution, I'll open things up and allow the students to work together on the remaining problems. Any that they are unable to complete in class will become tonight's homework.
It's worth pointing out that the problems on the second page all lead to remainders of zero. These problems are intended to lay a groundwork for the Remainder Theorem.