i was solving the question from the book IIT FOUNDATION AND OLYMPIAD - X and i was solving the problems of polynomials-III. so on the page number 136, there is a question (question 17) given below:

The remainder when $x$^100 is divided by $x^2-3x+2$ is:

a) $(2$^100$-1)x + (-2$^100$ +2) $

b) $(2$^100$+1)x + (-2$^100$ -2) $

c) $(2$^100$-1)x + (-2$^100$ -2) $

d) none

as far as i tried to find the remainder, i tried long division method but it was getting more and more complicated, then i used systematic method of division but i can't get the corret option what is the correct option. please explain me how did you find the remainder. thanks

and yes its answer is option (a)


Hint Write $x^{100}= (x^2-3x+2)q(x) + ax+b$. Now plug $x=1$ and $x=2$ to find $a$ and $b$.

  • $\begingroup$ what will i get after finding the value of a and b ? $\endgroup$ – anni Jun 21 '14 at 15:31
  • 1
    $\begingroup$ @anni, the remainder is $ax+b$. $\endgroup$ – lhf Jun 21 '14 at 15:31
  • $\begingroup$ okey i got it... thanks $\endgroup$ – anni Jun 21 '14 at 15:32

Hint $\,\ f\equiv g\ \ ({\rm mod} (x\!-\!1)(x\!-\!2))\ \iff x\!-\!1,x\!-\!2\mid f\!-\!g\ \iff f(1) = g(1),\ f(2) = g(2)$

  • $\begingroup$ should it go for none option ? $\endgroup$ – anni Jun 21 '14 at 15:28
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    $\begingroup$ @anni By above, with $\,f(x) = x^{100},\,$ an answer $\,g(x)\,$ is correct $\iff g(1) = 1,\ g(2) = 2^{100}.\ \ $ $\endgroup$ – Bill Dubuque Jun 21 '14 at 15:34

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