In what ways can educators introduce polynomials in grades 7 to 9?

Question

Q: Is there a way we can teach polynomials, avoiding the "watch me do it & now you do it" training method, that will allow students to anticipate and predict the existence or formulation of polynomials? (as it applies to, say, middle school mathematics)

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There are some properties of polynomials that may not be fully realized or proved at the grade 7 - 9 level. This is fine. I am more concerned about the introduction of polynomials to students. What I want to avoid is the particularly robotic and soulless routine of (some) math teaching which goes something like this:

1. here is a thing that's unfamiliar or you've never seen before
2. here is what we call some of the parts of this thing
3. here are some things you can do with them
4. now watch me do it and you do it until you get it right

Answer 1

One of the things I discovered with polynomials when working with them as a high-school student is that they generalize the idea of a place value system. This is a very cogent way to introduce the subject to middle-school students, who have probably been exposed to working in multiple bases.

For example, $11$ times $11$ is $121$ in every base except binary where the 2 in the middle gets carried out. This isn't a coincidence — it's because $(x + 1)(x + 1) = x^2 + 2x + 1$. Same with $12 \times 13 = 156$ for bases 7 and higher — it's just saying that $(x + 2)(x + 3) = x^2 + 5x + 6$, and evaluating it for $x = 10$.

So what you could do is to do a series of classes about working in different bases, and then ask — what if we work in a base with an unknown number? What happens then? Can we make arithmetic statements that work in all bases?

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If you use this place value paradigm, teaching polynomial arithmetic becomes easy as well — it's simply the algorithms they're used to, but you never carry anything like you would in a positional number system with a fixed base.

To multiply, say, $2x + 3$ by $x + 6$, you simply treat it as multiplying $23$ by $16$ in "base $x$", in which case you get $2\ 15\ 18$ as the "digits", or $2x^2 + 15x + 18$ in standard notation. They can verify this for themselves by plugging $x = 10$ and seeing that $23 \times 16 = 368$, which you would expect if you were to carry everything out. They can also plug in $x = 100$ to see that $203 \times 106 = 21518$, and that the carrying only happens when $x = 10$ because the digits are all scrunched together.

It will also help explain why certain mental tricks with multiplication work, such as the trick where you multiply a 2-digit number by $99$ by subtracting 1 from it, then appending the new number's difference from $99$ to the end (e.g. $43 \times 99 = 4257$). It works because multiplying a monomial $ax + b$ by $x^2 - 1$ gives you $ax^3 + bx^2 - ax - b$, which you can rewrite as $(ax + b - 1)x^2 + x^2 - ax - b$, and then $(ax + b - 1)x^2 + (x^2 - 1) - (ax + b - 1)$.

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Afterwards, you can discuss further differences between polynomials and regular place values, such as what happens when you use fractional coefficients. Why can't you represent, say, $\frac12 x^2$ as a polynomial with lower degree? What happens when you divide 100 by 2 in several different bases? Do they ever return the same thing, or even something similar?

Answer 2

What the OP left aside was:

Who cares? Why would we ever need to manipulate polynomials?

Permit me to answer this unasked question. Every^* piece of text you see rendered in PDF on your laptop screen, or printed out on a printer, draws each letter either with Adobe Postscript cubic Bézier polynomials, or Apple/Microsoft's TrueType, which uses quadratic Bézier polynomials.

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          ^{(Images from Mark Kilgard's presentaion. The dots are control points for the Bézier curves.)}

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We can hardly get through a day without seeing polynomials in action! :-)

As an advanced tanget, one could discuss the advantages/disadvantages of cubic vs. quadratic Bézier curves: continuity, control points, evaluation complexity, etc.

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^*Nearly every.

Answer 3

I like Joe Z's answer. It's how I introduce polynomials as well. In the interest of other approaches and since the idea is to introduce students to polynomials, here is another thought:

Consider a geometric approach and start with squares and cubes. If we have a square with side length $3$ units, then the area is $3^{2} = 9 \mbox{ units}^{2}$. A cube with side length $3$ would have volume $3^{3} = 27 \mbox{ units}^{3}$. A next step would be to use $x$ in place of the side length to obtain $x^{2}$ and $x^{3}$ for the area and volume, respectively.

Now, we can start to tease out some rules about polynomials if we keep the units hanging around. For example, we could ask what in the world it would mean to have $x^{3} + x^{2}$. How can we add volume with area? We can't in any sensible way, and hence we must maintain $x^{3} + x^{2}$, or analagously, $27 \mbox{ units}^{3} + 9 \mbox{ units}^{2}$ --- these units don't mix.

On the other hand, if I had two squares, I could join them and have their areas add to $2x^{2}$. Or have two cubes and have their volumes add to $2x^{3}$. All the while, we would carry our units along. Then we could really do some mind bending with some computation. Ask the question, "If I had three squares, each with side length three, what would my total area be?" The answer would be $27 \mbox{ units}^{2}$. Then we compare this result with one cube of side length three. The only difference is that, while in value, both quantities are equal in dimensionality they are not --- apples and oranges.

Multiplying polynomials can be brought back to finding the area of a square or the volume of a sphere. $x$ is a polynomial of degree 1. Thus, $x\times x = x^{2}$ and $x\times x\times x = x^{3}$, but $x\times x\times x = x^{2} \times x = x^{3}$ or $x\times x\times x = x\times x^{2} = x^{3}$.

I haven't really had much occasion to offer this type of explanation, but I keep it in reserve in case the approach as outlined by Joe Z just doesn't resonate. One distinct "advantage", is that it gets students to think about dimension and units.

This allows us to bring the conversation back to simple numerical substitution. What if we had $x^{3} + x^{2}$ and were asked to substitute $x = 3$? Why is the answer $27 + 9 = 36$ a valid answer, but in the case of our square and cube, we were forced to leave it as the nonsensical $27\mbox { units}^{3} + 9\mbox{ units}^{2}$? The answer is that $x = 3$ is dimensionless, whilst $x = 3 \mbox{ units}$ is not.