Homework

[Suzaku]

So now x^3+5x^2+5x+1=(x+1)(x^2+4x+1)

And therefore (x^4+4x^3-4x-1)/(X-1)=the above.

And what about those border values?

[Neb]

Never mind the none-PD stuff, I figured that out.

Now, I need a thing shortened: (X^3+3X-7)/(X^2-1)

It's too long to divide without making an awful mess of left-over numbers.

[Night Stalker]

Originally posted by Neb: Never mind the none-PD stuff, I figured that out. Now, I need a thing shortened: (X^3+3X-7)/(X^2-1) It's too long to divide without making an awful mess of left-over numbers.

No it's not. It's still the same thing. Just remember to use the filler terms. You need fillers in both terms this time but the process is the same.

Remember ....

Originally posted by Night Stalker: Whenever doing polynomial div and there is a missing term (x^2 in this case) you need to add a 0 filler. Start from the highest exp and count down left to right, there should be no break in numbers. ex: x^5 + x^3 + 1 needs to be rewritten as x^5+0x^4+x^3+0x^2+0x+1

This stuff ain't that bad .... you'll get .... ! [img]graemlins/blueblink.gif[/img]

[Neb]

I know that I need filler. But I don't think that I can reduce it with the filler there....

[LennonCook]

Actually, you don`t need the filler. It can help a little, but can also confuse matters.
To reduce a fraction, divide the top by the bottom; in this case
x^3+3x+7=(x^2-1)(x-1)+(3x+6)