Area models for multiplication throughout the K-12 curriculum

Let’s take a look at area models, shall we?

My thesis today is that area models should be ubiquitous across the entire curriculum because mathematics is a sense making discipline. As math educators, we ought to encourage our students to take every opportunity to visualize their mathematics in an effort to illuminate, explain, prove, and bring intuition.

So let’s take a walk through the K-12 math curriculum and highlight the use of area models as they might apply to arithmeticalgebra, and calculus.

base-ten-blocks

Arithmetic

Students experience area models for the first time in elementary school as they work to visualize multi-digit multiplication. This can also be used for division as well, just running the logic in reverse–that is, seeking an unknown “side length” rather than an unknown area. And Base Ten Blocks can be used to help students understand the building blocks of our number system.

Here’s how you might work out 27\times 54:

27\times 54 = (20+7)(50+4)=(20)(50)+(20)(4)+(7)(50)+(7)(4)

area-model-multiplication

27\times 54=1000+80+350+28=1458

The advantage of using a visual model like this is that you can easily see your calculation and explain why constituent calculations, taken together, faithfully produce the desired result. If you do a “man on the street” interview with most users or purveyors of the standard algorithm, you would almost certainly not get crystal clear explanations for why it produces results. For a further discussion of area models for multi-digit multiplication, see this article, or read Jo Boaler’s now famous book Mathematical Mindsets.

Algebra

In middle school, as students first encounter algebra, they may use area models to support their algebraic reasoning around multiplying polynomials. And in an Algebra 2 course they may learn about polynomial division and support their thinking using an area model in the same way they used area models to do division in elementary school. Here Algebra Tiles can be used as physical manipulatives to support student learning.

Here’s how you might work out (x+4)(2x+3):

(x+4)(2x+3)=(x)(2x)+(x)(3)+(4)(2x)+(4)(3)

area-model-polynomials

(x+4)(2x+3)=2x^2+3x+8x+12=2x^2+11x+12

Notice also that if you let x=10, you obtain the following result from arithmetic:

14\times 23 = 200+110+12=322

The Common Core places special emphasis on making such connections. I agree with this effort, even though I can also commiserate with fellow math teachers who say things like, “My Precalculus students still use the box method for multiplying polynomials!” We definitely want to move our students toward fluency, but perhaps we should wait for them to realize that they don’t need their visual models. Eventually most students figure out on their own that it would be more efficient to do without the models.

Calculus

Later in high school, as students first study calculus, area models can be used to bring understanding to the Product Rule–a result that is often memorized without any understanding. Even the usual “textbook proof” justifies but does not illuminate.

Here’s an informal proof of the Product Rule using an area model:

The “change in” the quantity L\cdot W can be thought of as the change in the area of a rectangle with side lengths L and W. That is, let A=LW. As we change L and W by amounts \Delta L and \Delta W, we are wondering how the overall area changes (that is, what is \Delta A?).

If the side length L increases by \Delta L, the new side length is L+\Delta L. Similarly, the width is now W+\Delta W. It follows that the new area is:

A+\Delta A=(L+\Delta L)(W+\Delta W)=LW+L\Delta W+W\Delta L+\Delta L\Delta W

area-model-product-rule

Keeping in mind that A=LW, we can subtract this quantity from both sides to obtain:

\Delta A=L\Delta W+W\Delta L+\Delta L\Delta W

Dividing through by \Delta x gives:

\frac{\Delta A}{\Delta x}=L\cdot\frac{\Delta W}{\Delta x}+W\cdot\frac{\Delta L}{\Delta x}+\frac{\Delta L}{\Delta x} \frac{\Delta W}{\Delta x} \Delta x

And taking limits as \Delta x\to 0 gives the desired result:

\frac{dA}{dx}=L\cdot\frac{dW}{dx}+W\cdot\frac{dL}{dx}

Conclusion

If you’re like me, you once looked down on area models as being for those who can’t handle the “real” algebra. But if we take that view, there’s a lot of sense-making that we’re missing out on. Area models are an important tool in our tool belt for bringing clarity and connections to our math students.

Okay, so last question: Base Ten Blocks exist, and Algebra Tiles exist. What do you think? Shall we manufacture and sell Calculus DX Tiles © ? 🙂

How do you expand √(a+b)?

This is a question that was recently asked on Quora:

it’s easy to expand
(a+b)^2 = a^2+2ab+b^2 or
(a+b)^3=a^3+3ab^2+3a^2b+b^3
or some other (a+b)^n but what about (a+b)^{1/2} aka. \sqrt{a+b}

Here’s my answer:

Just have Wolfram|Alpha do it for you :-).

But if you were on a desert island without access to Wolfram Alpha, here’s how you might think it through:

Are you already comfortable with the Binomial Theorem? Here it is again, but stated in a particular way that I think we’ll like.

\left(x+1\right)^r=1+rx+\frac{r(r-1)}{2!}x^2+\frac{r(r-1)(r-2)}{3!}x^3+\cdots

Look at it and make sure you understand it, and verify that it really is equivalent to the formulation of the Binomial Theorem you know.

Now, for the big trick. It turns out the above statement holds true not for just r=1,2,3,\ldots but for all real r. The only catch is that this often results in an infinite series. (These series results can also be obtained by Taylor expansion.)

In particular, it works for r=1/2:

\left(x+1\right)^{1/2}=1+\frac{1}{2}x+\frac{1/2(1/2-1)}{2!}x^2+\cdots

\left(x+1\right)^{1/2}=1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3+\cdots

Now, rewriting your original expression (a + b)^{\frac{1}{2}} as \sqrt{b}\left(a/b+1\right)^{1/2} gives

\sqrt{b}\left(1+\frac{1}{2}\left(\frac{a}{b}\right)-\frac{1}{8}\left(\frac{a}{b}\right)^2+\frac{1}{16}\left(\frac{a}{b}\right)^3+\cdots\right)

=\sqrt{b}+\frac{a}{2\sqrt{b}}-\frac{a^2}{8b^{3/2}}+\frac{a^3}{16b^{5/2}}+\cdots

which is the same result Wolfram Alpha will spit back.

Hope that helps!

Cubic polynomials and tangent lines

Just read an article in the most recent NCTM Mathematics Teacher magazine called “Students’ Exploratory Thinking about a Nonroutine Calculus Task” by Keith Nabb. I really, really enjoyed this article. Maybe for some this isn’t new, but I didn’t know this fact:

Average two of the roots of a cubic polynomial. Draw a tangent line to the cubic at this point. Did you know it will always pass through the third zero?? Incredible!

Here’s a nice site that I just googled that goes through one proof. However, the charm of the article mentioned above is that there are many interesting proofs that students came up with, some of which are more or less elegant (brute force algebra with CAS, Newton’s Method, just to name two of the four strategies mentioned in the article).

I wish I could give you the whole article, but you have to have an NCTM membership to see it. Here’s the link, but you’ll have to log in to actually see it.

Do Irrational Roots Come in Pairs? (Part 3)

continued from this post…

What polynomials can be solved?

Students are used to solving quadratic polynomials with the quadratic formula (if factoring techniques don’t work). And I mentioned in the previous post that Cardano gave us the very messy cubic formula. So it’s natural to ask, what polynomials can be solved?

The answer is that we can solve and get exact solutions for any polynomial up to degree four. This result is due to Ferrari and explained here (not for the faint of heart!!). It’s fun to give wolframalpha.com a fourth degree polynomial and see it go to work finding the exact zeros. Be sure to click on “Exact Form” to see the crazy nested radicals. Amazing what computers can do.

Fifth degree or higher degree polynomials can’t be solved by any particular formula or method. Interestingly, it’s not just that we haven’t discovered a method yet–it’s actually been proven impossible to solve a fifth degree polynomial. Évariste Galois is credited for this proof; he laid the foundation for Modern Algebra with some mathematics we now call Galois Theory. He proved that for any formula you write down that you claim solves the general 5th degree polynomial, we can construct a 5th degree polynomial that can’t be solved by your formula.

I think I have all my facts right. Pretty interesting stuff…and I don’t claim to fully understand it! Perhaps I’ll post more someday after some research.

 

[And thanks to Mr. Davis and Matthew Wright for inspiring me to post on these topics!!]

Do Irrational Roots Come in Pairs? (Part 2)

continued from this post…

Are all irrationals of the form a+b\sqrt{c}?

Consider, for instance, the simple third degree polynomial

g(x)=x^3-2

This function has one real root, x=\sqrt[3]{2}, and two nonreal roots. But notice that this root isn’t of the form a+b\sqrt{c}. There are lots of other irrationals that are not of this form. In fact, there are “more” irrationals not of this form than there are of this form (the set of irrationals of the form a+b\sqrt{c} are countable and the entire set of irrationals is uncountable). Here are just a few more that aren’t of the special form:

\pi, e, \sqrt[5]{7} and -\sqrt[3]{\frac{2}{3\left(9-\sqrt{69}\right)}}-\frac{\sqrt[3]{\frac{1}{2}\left(9-\sqrt{69}\right)}}{3^{2/3}}

These irrationals seem a bit more contrived. This is an example of where our intuition doesn’t match reality. In fact, most real numbers are impossible to describe at all. This is very hard to believe, even though it’s true.  So we, necessarily, don’t talk about most numbers! On another note, \pi and e will never ever be roots of polynomials (which is why we call them transcendental).

Another example

Here’s another example of a polynomial with one irrational root that came up in our Precalculus homework this past week:

h(x)=x^3-x+1

This has only one real root. It’s an irrational root, and so it must not be of the form a+b\sqrt{c}. In fact, the  one real zero of h(x) is the last irrational number in the list above. How do we find such a convoluted answer?

The answer is we use Cardano’s Method, which works for cubic equations (it would work particularly nicely on the above  polynomial). But for higher degree polynomials, we can only hope to attack it using various algebraic tools like Rational Root Theorem, Descartes Rule of Signs, good guessing, long division, substitutions,  factoring techniques, or other sneaky algebraic tricks. If algebraic techniques fail, all we can do is resort to approximation (usually using Newton’s Method). So, for the first polynomial we started with,

f(x)=-x^5+2x^4+7x^3+x^2-4x+1

has one irrational root, but I don’t know how to find it except by approximation: It’s approximately 3.83. And WolframAlpha doesn’t know either.

So there we have it, real polynomials with rational coefficients can have ONE irrational root. Let no one convince you otherwise! 🙂

(And here’s another great discussion of this topic.)