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# Station  13Division in any base

Here’s the division problem $276 \div 12$ we did earlier in a $1 \leftarrow 10$ machine. We see the answer $23$.
Stare at this picture for a moment – it will soon sneak back up on us.

### Question 1

Let’s now do the same division problem in another base. But the only tricky part is that I am not going to tell you which machine we are in!

We could be in a $1 \leftarrow 10$ machine again, I am just not going to say. Maybe it will be in a $1 \leftarrow 2$, or a $1 \leftarrow 4$ machine or a $1 \leftarrow 13$ machine.

Since I don’t seem to be telling which machine we’re in let’s call it an $1 \leftarrow x$ machine. (People always seem to use the letter $x$ to represent a quantity whose value they do not know.)

Try putting some dots in the rightmost box and make them explode.
You’ll see that it is quite annoying that I am not telling you which machine we’re in!

### Question 2

In a $1 \leftarrow 10$ machine the place-value of boxes were the powers of ten: $1, 10, 100, 1000, ...$.
In a $1 \leftarrow 2$ machine the place-value of boxes were the powers of two: $1, 2, 4, 8, 16, ...$.
In a $1 \leftarrow x$ machine the place-values of boxes must be the powers of $x$: $1,x,x{2},x{3}, ...$.

As a check:
If I tell you $x$ is ten in my mind, then we really are getting $1, 10, 100, 1000, ...$ and if I tell you instead $x$ really is two, then we are getting $1, 2, 4, 8, 16, ...$, and so on. So the $1 \leftarrow x$ machine is indeed representing ALL machines all at once.

Okay … without any warning, here’s a high-school algebra problem:
Compute $\left( 2x^{2}+7x+6\right) \div \left( x+2\right)$.

### Question 3

This high school algebra problem is identical to a grade school arithmetic problem!

What’s going on?

Suppose I told you that $x$ really was $10$ in my head all along. Then $2x^{2} + 7x + 6$ is the number $2 \times 100 + 7 \times 10 + 6$, which is $276$. And $x + 2$ is the number $10 + 2$, that is, $12$. And so we computed $276 \div 12$. We got the answer $2x + 3$, which is $2 \times 10 + 3 = 23$, if I am indeed now telling you that $x$ is $10$.

So we did just repeat a grade-school arithmetic problem!

Aside: By the way, if I tell you that $x$ was instead $2$, then

$\left( 2x^{2}+7x+6\right) = 2 \times 4 + 7 \times 2 + 6$, which is $28$,
$x + 2 = 2 + 2$, which is $4$,
and,
$2x + 3 = 2 x 2 + 3$, which is $7$

Ansd also we just computed $28 \div 4 = 7$, which is correct.

Doing division in an $1 \leftarrow x$ machine is really doing an infinite number of division problems all in one hit.
Whoa!

Compute $\left( 2x{3}+5x{2}+5x+6\right) \div \left( x+2\right)$ in the $1 \leftarrow x$ machine.

### Question 4

Use the following questions to master the $1 \leftarrow x$ machine.

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

### Question 5

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

### Question 6

Here’s a polynomial division problem written in fraction notation. Can you do it?

$\dfrac {x{4}+2x{3}+4x^{2}+6x+3} {x^{2}+3}$

### Question 7

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

Let's go to the next station!

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