Algebraic Expressions Really Are Numbers! (Or is it the other way around?)

## Algebraic Expressions Really Are Numbers! (Or is it the other way around?)

So, you know when they write something like $x+4=7$, and you go ...

Yeah, $x$ is $3$. Easy. So what?

Well, d'you ever think to read these equations like someone is stating a fact? Like "California is west of Pennsylvania."

I mean, when, let's say that I said it out, like: "$X$ is a number, and $x+4$ is a different number that's 4 more than $x$, and that different number, $x+4$, is $7$, not just equal to $7$, but really is the number $7$."

Now I'm really not following you! What do you mean that "$x+4$ is not just equal to $7$, but really is $7$"? They sound the same to me.

Yeah, but it's not, really. I mean, if I say that someone looks like you and acts like you, and has your name, that's possible right?

Unlikely, but possible, I guess. Like a clone or something?

Right, maybe your twin. But they aren't you, right?

Sure, obviously.

Well, numbers don't have clones like that. Seven is just seven. Every expression that refers to seven, when I write it out in letters S-E-V-E-N, or write the number, $7$, or write it like $3+4$, they are all exactly the same thing -- the number that's one more than $6$ and one less than $8$.

Where are you going with this?

Okay, so when I say $x+4=6$, and if there's only one thing that is $6$, then when I say that $x+4=6$ I'm saying that the expression $x+4$ is just another way of writing $6$. Just like writing it out in letters S-I-X, and so on."

You've really totally lost me.

Okay, let's talk about something else.

Good idea!

You remember long division?

Sure. It's been a while, but I think I can manage it.

Great, so first multiply $43$ by $62$.

You want me to use long multiplication?

A calculator is fine, if you need it.

$2666$

Great, so now divide 2666 by $43$.

Um, $62$? Is this a trick question?

No, I mean divide it using log division.

Oh, okay. I think I can manage that.

$$\require{enclose} \begin{array}{rll} 62 && \hbox{(Explanations)} \\[-3pt] 43 \enclose{longdiv}{2666}\kern-.2ex \\[-3pt] \underline{258\phantom{0}} && \hbox{(43 \times 6 = 258)} \\[-3pt] 8\phantom{0} && \hbox{(266 - 258 = 8 )} \\[-3pt] {\phantom{0}86} && \hbox{(Bring down the 6)} \\[-3pt] \phantom{0}\underline{86} && \hbox{(43 \times 2 = 86)} \\[-3pt] {\phantom{00}0} && \hbox{(86 - 86 = 0)} \\[-3pt] \phantom{00}Done! \end{array}$$

See: $62$!

Great. So now do this for me: $(4x+3)\times(6x+2)$.

You mean just FOIL it? Um ... $24x^2+18x+8x+6$ so, uh ... $24x^2+26x+6$

Excellent, so Great. So what's $24x^2+26x+6$ divided by $(4x+3)$?

Wait; What?! I don't know how to do that!

Sure you do. You just multiplied them up there, so just undo the multiplication!

Oh. I see. Sure, um I guess it's $(6x+2)$, right?

Exactly! But, now to it by long dibision!

Hunh? Now I have no idea what you're talking about.

No, really. It works exactly like long divison of numbers! Here, I'll show you:

$$\require{enclose} \begin{array}{rll} (4x+3) \enclose{longdiv}{24x^2+26x+6}\kern-.2ex \\[-3pt] \end{array}$$

I have no idea what to do with that!

Just do what you usually do. Ask what you have to multiply $(4x+3)$ by in order to be able to subtract it out from the head of $24x^2+26x+6$

Um, $6$?

That would make $24x+18$, but it looks like you need an $x^2$...

$6x$ then? That would make it $(24x^2+18x)$, but the 18 doesn't match up.

No worries. Let's do exactly what we would do in a long division: Put $(24x^2+18x)$ under the head of the long experssion, and write the $6x$ where it would normally go, over the end of the subexpression, like this:

$$\require{enclose} \begin{array}{rll} {6x\phantom{0@0}} && \\[-3pt] (4x+3) \enclose{longdiv}{24x^2+26x+6}\kern-.2ex \\[-3pt] 24x^2+18x\phantom{00@} && \\[-3pt] \end{array}$$

Okay, now we just subtract in the normal way we would in long division:

$$\require{enclose} \begin{array}{rll} {6x\phantom{0@0}} && \\[-3pt] (4x+3) \enclose{longdiv}{24x^2+26x+6}\kern-.2ex \\[-3pt] \underline{24x^2+18x}\phantom{00@} && \\[-3pt] 8x\phantom{00@} && \\[-3pt] \end{array}$$

Now we'll bring down the next term, $+6$, in the normal way:

$$\require{enclose} \begin{array}{rll} {6x\phantom{0@0}} && \\[-3pt] (4x+3) \enclose{longdiv}{24x^2+26x+6}\kern-.2ex \\[-3pt] \underline{24x^2+18x}\phantom{00@} && \\[-3pt] 8x+6 && \\[-3pt] \end{array}$$

Oh oh! I see, and $4x+3$ times $2$ is $8x+6$, and we put the $+2$ on top, at the end!

$$\require{enclose} \begin{array}{rll} {6x+2} && \\[-3pt] (4x+3) \enclose{longdiv}{24x^2+26x+6}\kern-.2ex \\[-3pt] \underline{24x^2+18x}\phantom{00@} && \\[-3pt] 8x+6 && \\[-3pt] \underline{8x+6} && \\[-3pt] {\phantom{00}0} && \\[-3pt] \phantom{00}Done! \end{array}$$
$$\require{enclose} \begin{array}{rll} 5x^2-3x+4 && \hbox{(Explanations)} \\[-3pt] (x-3) \enclose{longdiv}{16x^3-4x^2-3x-23}\kern-.2ex \\[-3pt] \underline{4\phantom{00}} && \hbox{(4 \times 1 = 4)} \\[-3pt] 10\phantom{0} && \hbox{(5 - 4 = 1)} \\[-3pt] \underline{\phantom{0}8\phantom{0}} && \hbox{(4 \times 2 = 8)} \\[-3pt] \phantom{0}20 && \hbox{(10 - 8 = 2)} \\[-3pt] \underline{\phantom{0}20} && \hbox{(4 \times 5 = 20)} \\[-3pt] \phantom{00}0 \end{array}$$ $$\require{enclose} \begin{array}{rll} 5x^2-3x+4 && \hbox{(Explanations)} \\[-3pt] (x-3) \enclose{longdiv}{16x^3-4x^2-3x-23}\kern-.2ex \\[-3pt] \underline{4\phantom{00}} && \hbox{(4 \times 1 = 4)} \\[-3pt] 10\phantom{0} && \hbox{(5 - 4 = 1)} \\[-3pt] \underline{\phantom{0}8\phantom{0}} && \hbox{(4 \times 2 = 8)} \\[-3pt] \phantom{0}20 && \hbox{(10 - 8 = 2)} \\[-3pt] \underline{\phantom{0}20} && \hbox{(4 \times 5 = 20)} \\[-3pt] \phantom{00}0 \end{array}$$