Finding the remainder when x^80 - 8x^30 + 9x^24 + 5x + 6 is divided by (x+1)

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duki
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Homework Statement



Find the remainder when (x^80 - 8x^30 + 9x^24 + 5x + 6) is divided by (x+1)

Homework Equations



The Attempt at a Solution



So I'm not really sure where to start. I tried starting by doing long polynomial division, but I get stuck. How do I start this?
 
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Yeah, you would get stuck doing the division. It's a long haul. But look, suppose you did do the division f(x)=(x^80 - 8x^30 + 9x^24 + 5x + 6) by (x+1)? Then you would get f(x)=q(x)*(x+1)+r, right? Where q(x) is the quotient and r is the remainder. What happens if you put x=(-1) into that?
 
Remainder Theorem:
If p(x) / (x – a) = q(x) with remainder r(x),

then p(x) = (x – a) q(x) + r(x).
 
Dick said:
Yeah, you would get stuck doing the division. It's a long haul. But look, suppose you did do the division f(x)=(x^80 - 8x^30 + 9x^24 + 5x + 6) by (x+1)? Then you would get f(x)=q(x)*(x+1)+r, right? Where q(x) is the quotient and r is the remainder. What happens if you put x=(-1) into that?

You get r ?
 
Well, yes. You get f(-1)=r. That's the remainder theorem. So what is r?
 
Sure. If you don't believe it make up a simpler example where you can actually do the long division and check that it works. It's good for you.
 
duki said:
Thanks. How did you know to use -1?

Look back at the problem. I'll give you three guesses. The first one had better be right.
 
What's a? If a=x+1 then x=a-1. You are onto the second guess.
 
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Dick said:
What's a? If a=x+1 then x=a-1. You are onto the second guess.

Well heck, I'm not sure. :O
I assume we're using -1 because of something to do with (x+1) = 0 or something?
 
Yes, if you had spent all day figuring out the q(x) in f(x)=q(x)*(x+1)+r by doing the horrible division, at the end of it all you could realize that you didn't need to find q(x) at all because if you put x=(-1) the q(x) disappears. That's the remainder theorem.
 
Sweets.
So if for example I was dividing by (x-4), I would use 4 instead of -1?
 
f(x)=q(x)*(x-4)+r. Sure, f(4)=r. You don't need to find q(x) before you know the remainder.