How do you divide #(x^101+x+1)/( x^2+1) #?

1 Answer
May 8, 2016

Quotient = #x^99-x^97+x^95-...+x^3-x#

Remainder: #1+2x#

Explanation:

Use #(X^n+1)/(X+1)=X^(n-1)-X^(n-2)+...+1#, n = 3, 5, 7,...

#=X^(n-1)-X^(n-2)+...+X-1 +2/(X+1),. n=2. 4, 6,...#.

Here, the second case is relevant.

The ratio is #(x((x^2)^50+1)+1)/(x^2+1)#

#=x(((x^2)^50+1)/(x^2+1))+(1/(x^2+1))#

#=x(((x^2)^49-(x^2)^48+..+(x^2)-1)+2/(x^2+1))+1/(x^2+1)#

#=x^99-x^97+..+x^3-x+(1+2x)/(x^2+1)#

I did not correctly remember the relevant formula. It was like remembering the spelling of 'caret' as 'carat'.. In the present edition, I have now made self-correction for the correct answer.. I am sorry for the trouble.

I agree that long-hand-division easily gets the quotient and remainder. Yet, the formulas given by me are relevant.