Nov 072014

This post is a branch of yesterday’s post on polynomials. As were are teaching this year, I am giving them some tough, torturous problems that are about incorporating all the math from previous units into one problem each time.

Yes, I want every problem to be a review of the previous year’s materials. And it is working so far. Another teacher and were looking at this problem set at lunch and we were discussing whether or not we had to have pairs of parenthesis that were conjugates.


I mean, I did give them pairs that were conjugates in each of these, but do I have to?

No. And not just no, but why should I?

Let’s think of a polynomial differently. A polynomials is nothing more that a series of “lines” that are multiplied. Those “lines” can be real lines  (x+2) or irrational lines (x+2sqrt(3)) or even imaginary “lines” (x+2i). When we take these “lines” and multiply them together we get a polynomial function.

cubic1 or cubic2

or even! cubic3. Yes, I had to do this in the Nspire software because this last graph breaks Desmos. Desmos cannot graph the imaginary numbers of the intercept form of a polynomial. #sadness

But notice that in each of these graphs I used the conjugates. Did I have to?



Not at all. Here is a polynomial function, perfectly formed, that is composed of three lines multiplied together. Can I do it with three different imaginary numbers? No. That is where we MUST use conjugates. Only in that one, special case.

So the question is why do some teachers think that conjugates are required in ALL cases of polynomials? The reason is very simple. Only when we have conjugates can we form a standard form function that can be solved with the quadratic formula.

That’s it. That is the only reason.

I was pleasantly surprised to find this series of old posts by Mr. Chase this week as well:

Do Irrational Roots come in Pairs Part 1

Do Irrational Roots come in Pairs Part 2

Do Irrational Roots come in Pairs Part 3

He approaches it from the standard form side, but I think it is far easier to see why it is perfectly okay from the intercept form side of a polynomial.

The real issue is do you have to have conjugates of imaginary numbers in a polynomial. I think, and I do emphasize the THINK, that we do. If we have an imaginary number in the standard form polynomial then we can not graph it on the real plane.  I have been trying to think of a work around (reminiscent of what I did with the quadratics here and here.) Still wrapping my head around it. Not sure if I can see a way out.


How does this relate to my class? Well, my learners are doing problems like the fourth picture. That counts.

  3 Responses to “Building & solving polynomials”

  1. If you want 3√2 as a root, AND you want your polynomial to have *rational* coefficients, you must also have -3√2 as a root. This is completely analogous to nonreal roots. If you want 3i to be a root AND you want your polynomial to have *real* coefficients, then -3i must also be a root.

    If you drop any of those requirements, you can have a bit more freedom with your roots. There’s nothing inherently wrong with irrational or nonreal coefficients on a polynomial–such a polynomial would just be complex-valued. In Complex Analysis we often plot a function with two coordinate grids (the w and z planes) to show pre-image and image of a function/transformation. Color or animations can also help visualize Complex functions.

    I love this stuff!

    PS: in one of your linked posts you say “determinant,” but I think you mean “discriminant.”

  2. Doh! Thank you John. I will hunt that down and fix it.

    I love this stuff too. To think about why we say what we do as teachers, and the limitations we place on learners because of it is so important. And, we discover really cool math when we shed the misconceptions and preconceptions we have. Thank you!

  3. […] in point: Glenn’s (@gwaddellnvhs) obsession about how everything is just made up of lines (It’s somehow comforting to know that Dylan struggled with this idea as […]

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