if x^y means x to the power y then
f(x)=1+x+x^2+....+x^6 then the reminder when f(x^7) is devided by f(x)

• Given that f(x7)=1+x7+(x7)2 + ....+ (x7)6 = 1+x7+x14+....+x42
  We will rewrite the above equation, f(x7)=1+(x7−1)+(x14−1)+... + (x42−1)+6
  We know that x7−1=(x−1)(x6+x5+...1)
  Now It is clear that x7−1 is exactly divisible by f(x).
  Also x14−1=(x7)2−12 and x7−1 is a factor of this expression. (∵xn−an is always divisible by x−a
  Similarly, we write x21−1=(x7)3−13, x28−1=(x7)4−14....
  So remainder = 1 + 6 = 7
• 9 years agoHelpfull: Yes(7) No(1)
• ANS=0.
  let x=1,
  f(1)=1 + 1^2 + 1^3 + 1^4 + 1^5 + 1^6.
  f(1)=7.
  now, f (x^7) = f (1^7) = f (1) => 7
  f (x^7) / f (x) = f(1) / f(1) =0(remainder).
• 9 years agoHelpfull: Yes(1) No(8)