Due to the Remainder theorem, the remainder of division is the value of the polynomial p(x) at x= 1.
So, substitute x= 1 into the polynomial and calculate the obtained alternate sum.
Since the degree 2002 is an even number, the number of " + 1 " addends will be one more
than the number of " -1 " addends, so the alternate sum is 1 (one).
ANSWER. The remainder is 1.
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The remainder theorem says:
Divide p(x) over (x-k). You will get some quotient q(x) and a remainder of p(k)
Consider these smaller degree polynomials
f(x) = x^2 - x + 1
g(x) = x^4 - x^3 + x^2 - x + 1
h(x) = x^6 - x^5 + x^4 - x^3 + x^2 - x + 1
each of which is even degree
The signs alternate between plus and minus from term to term.
We'll divide each of those functions over (x-1). So we'll plug in x = 1 into each function.
f(1) = 1^2-1+1 = 1-1+1 = 0+1 = 1
g(1) = 1^4-1^3+1^2-1+1 = (1-1)+(1-1)+1 = 0+0+1 = 1
h(1) = 1^6-1^5+1^4-1^3+1^2-1+1 = (1-1)+(1-1)+(1-1)+1 = 0+0+0+1 = 1
Each time we get 1 as a result.
Notice the +1 and -1 terms pairing up to cancel out
The +1 at the very end is the lone survivor after all those cancellations.
Extend this idea out to a 2022 degree polynomial and we'll have those same cancellations going on, and end up with a remainder of 1
p(x) = x^2022-x^2021+x^2020+...-x^3+x^2-x+1
p(1) = 1^2022-1^2021+1^2020+...-1^3+1^2-1+1
p(1) = (1-1) + (1-1) + ... + (1-1) + 1
p(1) = (0) + (0) + ... + (0) + 1
p(1) = 1
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