find the remainder when x51+51 is divided by (x+1).

Question

anonymous · Answer

is that:
$$x^{51}+51$$
?

anonymous · Answer

yes

anonymous · Answer

(which btw, is ridiculous, whoever asked you to do this needs to be punched in the face.)

anonymous · Answer

why?

anonymous · Answer

because its to the 51st power.  Thats waaaaaay too high for any normal type of problem.
$$x^4 +3$$
is normal.
a polynomial of degree 51 is retarded >.< anywhos, that just my personal feelings on the subject, now for the solution lol

anonymous · Answer

ok

anonymous · Answer

Do you know about synthetic division, or does that term sound familiar to you?

anonymous · Answer

srry.i dont noe dat

anonymous · Answer

alright, thats fine, theres is still one more method, i will have to write it out on paper and post it though. One sec.

anonymous · Answer

ok

anonymous · Answer

if you dont mind me asking, what lvl class is this? algebra II? Calculus?

anonymous · Answer

nvm, its not that rough of a problem.  Basically you do this:
$$x^51+51 = (x+1)p(x)+r$$
where p(x) is the quotient, and r is the remainder.  The (x + 1) is there because we are dividing by (x + 1).  If we plug in the value -1 for x we get:
$$(-1)^{51}+51 = (-1+1)p(x)+r \Rightarrow -1+51 = r \Rightarrow r = 50$$
so the remainder is 50.

anonymous · Answer

oops, that should be a 
$$x^{51}$$
in the first equation line.