First of all, the A, B, and C are different from my post above.  The general form of a second degree polynomial (ie, one with it's highest term being x²) is Ax²+Bx+C.  In the case above, A=1, B=-1, C=-12.

Now, to multiply something of the form (d*x+p)(e*x+q), we use a process called FOIL.  This stands for First, Outer, Inner, Last.  This means you first multiply the first terms together, which is d*x and e*x in this case, to get d*e*x².  The Outer refers to the first term, and the last term being multiplied together.  In this case, d*x and q are multiplied together to get e*x*q.  Inner refers to the two inside terms, e*x and p. Multiplying them together gives px.  Also note that p and q are usually constants, so you can combine them by (e*p+d*q)x.  Lastly, or Last as it were, we have p and q being multiplied together, to get pq.

Now, remember the A, B, and C in the Ax²+Bx+C.  To find the d, e, p, and q, that fits a certain A, B, and C, we have to figure these equations out:
A=d*e
B=e*p+d*q
C=p*q
Now, A is 1 in this case, so it simplifies to:
A=1
B=p+q
C=p*q
B is -1 and C is -12.  In this case, -4 and 3 work because -3*4=-12=C and -3+4=-1=B.

Does that all make sense?

((x²-x-12))/x)/(x/(x-4))


Okay, so let's start piece by piece and see where we go.

First of all, you can factor (x²-x-12) to (x-4)(x+3).  This gives ((x-4)(x+3))/x)/(x/(x-4)).

Now, remember how A/B is the same as A times (1/B)?  Well, the same thing works with A/(B/C), however, 1/(B/C) is the same as C/B.  So A/(B/C) is the same as A times (C/B).  This means that ((x-4)(x+3))/x) which is A in this case, divided by (x/(x-4)) which is B/C in this case, gives ((x-4)(x+3))/x) times ((x-4)/x).  Combining these gives ((x-4)(x-4)(x+3)/x²).

Now, I don't know exactly where you want to go with this, but that is pretty simplified there.  You could multiply the numerator terms together and get (x²-x-12)(x-4) or (x³-5x²-16x+48).

So (x³-5x²-16x+48)/x² is an answer.

EDIT: Also, if you set it equal to zero, you could disregard that x² and use (x-4)(x-4)(x+3)=0.  So you have x=4, 4, and -3.

@Graph; What do you mean faster in real life? The screenshot is 30fps

It seems it just is slow for displaying the .gif.  Can't wait to play!

Yeah, I say add them tentatively.  Like if you have X number of posts or something, otherwise just do a browser extension.

I voted to yes, add them, but see posts. ie, this post.

Is that a right-angle symbol in the top right triangle?  And is that entire line 8 or what?

Photoshop works on video now? When did this happen?

The last version I checked, you could edit the frames in a video.  Don't know how much this has changed, though, and that was CS4 or something.

Quote from: Ashbad on April 08, 2011, 06:26:48 am

.org. Is good to use too, same with .reallocate

And there are others, but most people don't use them much, and still they're mostly used for z80 assembly development.

Ashbad, I can't find the meaning of .reallocate anywhere. What does that do? And the .org was my bad there. But actually unlike on the z80, most people will not use .org because most SH3 code is location independent and I've also been pushing for shell-cross compatibility for all prizm programs.

Quote from: graphmastur on April 08, 2011, 07:32:48 am

; actually isn't a new-line character.  And there are two new line characters, one is 1 byte, one is two.  One is a line-break and feed I believe, the other is just a line break.

The 1 byte one is 0x0A, the 2 byte one is 0x0D0A

I was just mentioning how in C/C++ you can start a new line using a ; but that is already reserved in asm for indicating a comment. What I'm looking for is the character allowing you to use multiple commands in one line.

Oh, probably \.  That's what it usually is, IIRC.

Indeed, but that would be so un-TI of TI

It was un-TI of them to say that you could downgrade in the link before downloading too...