Exploding Dots

1.7 Arithmetic in an \(1 \leftarrow x\) Machine

Lesson materials located below the video overview.

Here’s the division problem \(384 \div 12\) worked out in a \(1 \leftarrow 10\) machine. We see \(384 \div 12 = 32\).

ex1701

But I want to now break away from our human predilection for ten-ness. I would like to now work in a different machine, maybe the \(1 \leftarrow 2\) machine we introduced right at the start of the course, or maybe the \(1\leftarrow 3\) machine, or maybe a \(1 \leftarrow 7\) machine, or maybe a \(1 \leftarrow 207\) machine.

 

The point is that I refuse to tell you which machine we are in. And since we don’t know which machine we’ll be playing in, let’s just call it an \( 1 \leftarrow x\) machine. Maybe \(x\) is \(10\), or it is \(2\), or it is \(82\). I am just not telling.

 

Now for a \(1 \leftarrow 10\) system the boxes have values the powers of ten: \(1\), \(10\), \(100\), \(1000\), \(10000\), ….
In a \(1 \leftarrow 2\) system, the powers of two: \(1\), \(2\), \(4\), \(8\), \(16\), …, and so on.

 

In my system, whatever it is, we must have powers of \(x\):

ex1702

So in this system, when I write “\(384\)” I actually mean:

\(3x^{2} + 8x + 4\).

ex1706

(Of course, if I now tell you that I was actually thinking \(x\) was \(10\) all along, then this really does correspond to the number three-hundred and eighty four. If, however, I was thinking that \(x\) is \(3\), say, then this is \(3 \times 9 + 8 \times 3  + 4 \times 1 = 55\).)

 

Now let’s work out:

\(\left( 3x^{2}+8x+4 \right) \div \left( x + 2 \right) \).

It’s a déjà vu experience!

ex1707

We see that:

\(\left( 3x^{2}+8x+4 \right) \div \left( x + 2 \right) = 3x+2\).

 

Thus the division of polynomials can be regarded as a elementary school long division!

 

Comment: If \(x\) is indeed \(10\) we have just computed \(384 \div 12 = 32\). If \(x\) is \(3\), we’ve just computed \(55 \div 5 = 11\). (Do you see why?)

 

Question 38:  Show that \(\left( 2x^{3} + 5x^{2} + 5x + 6\right) \div \left( x + 2 \right) \) equals \( 2x^{2} + x + 3\). What ordinary division problem have you just computed if I tell you that \(x\) was \(10\) all along?

 

Question 39:  Compute the following:

a) \(\left( 2x^{4} + 3x^{3} + 5x^{2} + 4x + 1\right) \div \left( 2x + 1 \right) \)

 

b) \(\left( x^{4} + 3x^{3} + 6x^{2} + 5x + 3\right) \div \left( x^{2} + x + 1 \right) \)

 

 

Do this next question for sure. If you can handle it, then you can handle anything. (Well … almost anything! We have a concern coming up.)

 

Question 40: Is this division problem straightforward or is it problematic?

\( \dfrac{x^{4} + 2x^{3} + 4x^{2} + 6x + 3}{x^{2} + 3}  \)

 

(Here we wrote the division problem in fraction-like notation.)

 

Question 41: Show that \(\left( x^{4} + 4x^{3} + 6x^{2} + 4x + 1\right) \div \left( x + 1 \right) \) equals \(x^{3} + 3x^{2} + 3x + 1\) .

a) What is this saying for \(x = 10\)?

b) What is this saying for \(x = 2\)?

c) What is this saying for \(x\) equal to each of 3, 4, 5, 6, 7, 8, 9, and 11?

d) What is it saying for \(x = 0\)?

e) What is it saying for \(x = -1\)?

 

 

DASHING OUR HOPES!

So it looks like our dots and boxes method handles the traditional algebra of polynomials one learns in an algebra II course. Everything is ridiculously easy, at least conceptually, and makes complete perfect sense.

But I have perhaps been pulling the wool over your eyes.

 

TRY THIS NEXT PROBLEM ON YOUR OWN BEFORE CONTINUING WITH THIS COURSE
Can it be solved via our methods? Are you stuck? Is there an epiphany to be had?

 

\(\dfrac{x^{3}-3x+2}{x + 2}\)

 

Warning: It is tempting to unexplode dots for this problem. But remember, I am NOT telling you what value I have in mind for \(x\):  you don’t know how many dots to draw when you unexploded one – and that is mighty annoying!

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