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\n" ); document.write( "The remainder theorem says:
\n" ); document.write( "Divide p(x) over (x-k). You will get some quotient q(x) and a remainder of p(k)\r
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\n" ); document.write( "\n" ); document.write( "Consider these smaller degree polynomials
\n" ); document.write( "f(x) = x^2 - x + 1
\n" ); document.write( "g(x) = x^4 - x^3 + x^2 - x + 1
\n" ); document.write( "h(x) = x^6 - x^5 + x^4 - x^3 + x^2 - x + 1
\n" ); document.write( "each of which is even degree
\n" ); document.write( "The signs alternate between plus and minus from term to term.\r
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\n" ); document.write( "\n" ); document.write( "We'll divide each of those functions over (x-1). So we'll plug in x = 1 into each function.
\n" ); document.write( "f(1) = 1^2-1+1 = 1-1+1 = 0+1 = 1
\n" ); document.write( "g(1) = 1^4-1^3+1^2-1+1 = (1-1)+(1-1)+1 = 0+0+1 = 1
\n" ); document.write( "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
\n" ); document.write( "Each time we get 1 as a result.
\n" ); document.write( "Notice the +1 and -1 terms pairing up to cancel out
\n" ); document.write( "The +1 at the very end is the lone survivor after all those cancellations.\r
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\n" ); document.write( "\n" ); document.write( "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
\n" ); document.write( "p(x) = x^2022-x^2021+x^2020+...-x^3+x^2-x+1
\n" ); document.write( "p(1) = 1^2022-1^2021+1^2020+...-1^3+1^2-1+1
\n" ); document.write( "p(1) = (1-1) + (1-1) + ... + (1-1) + 1
\n" ); document.write( "p(1) = (0) + (0) + ... + (0) + 1
\n" ); document.write( "p(1) = 1\r
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\n" ); document.write( "\n" ); document.write( "Answer: 1
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