Let P(x)=1+x+x2+x3+x4+x5
What is the remainder when P(x12) is divided by P(x)??
(A)0 (B)6 (C)1+x (D)1+x+x2+x3+x4
11is the answer b.
my method is that p(-1)=0
now to find the remainder of p(x12)=g(x0)
you can simply find g(-1)=6(this is because if you factorize p(x) you find that its other roots are complex which cannot contribute to the remainder of g(x0) .also you can verify by factorizing g(x0) further.)
341Simply do this
P(x^{12}) =1+x^{12}+x^{24}+x^{36}+x^{48}+x^{60}
= =6+(x^{12}-1)+(x^{24}-1)+(x^{36}-1)+(x^{48}-1)+(x^{60}-1)
Each of the bracketed expressions is divisible by x^6-1 and hence by P(x)
341A more general discussion can be read at http://www.goiit.com/posts/list/algebra-question-56626.htm
1after prophet sir's #4 i dont think there's should be any kind of ambiguity..
