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Consider a function $f(x) = x^4+x^3+x^2+x+1$, where x is an integer, $x\gt 1$. What will be the remainder when $f(x^5)$ is divided by $f(x)$ ?

$f(x)=x^4+x^3+x^2+x+1$

$f(x^5)=x^{20}+x^{15}+x^{10}+x^5+1$

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4 Answers 4

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Rewrite the expression as $$(x^{20}-1) +(x^{15}-1)+(x^{10}-1)+(x^5-1)+1+4.$$ The expression $x^4+x^3+x^2+x+1$ divides the first four terms, since $x^5-1$ does. So the remainder is $5$.

The idea obviously generalizes.

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  • $\begingroup$ Nice solution André Nicolas..... $\endgroup$
    – juantheron
    Dec 31, 2013 at 4:30
  • $\begingroup$ can some one please explain this solution a little bit more as I am not able to understand it properly. $\endgroup$
    – Ganit
    Sep 6, 2021 at 8:46
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Let $x^{20}+x^{15}+x^{10}+x^5+1=(x^4+x^3+x^2+x+1)Q(x)+R(x)$.

$x^4+x^3+x^2+x+1=0$ has $4$ complex roots $a_1,a_2,a_3,a_4$. And these are also roots of $x^5=1$, so when $x=a1,a2,a3,a4$, the above equation becomes

$5=R(a_i)$ ($i=1,2,3,4$)

This is true when $R(x)=5$ for every $x$, and it is easy to show that a polynomial of degree $3$ like this is unique. So the remainder is just $5$ (constant).

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Observe that $f(x^5) = f(x)(x^{16}-x^{15}+2x^{11}-2x^{10}+3x^6-3x^5+4x-4) + 5$.

So the remainder will be $5$.

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$f_1(x)=x+1$

$f_1(x^5)=x^5+1$

$f_1(x^5)$, when divided by $f_1(x)$ leaves a remainder 0.

$f_2(x)=x^2+x+1$

$f_2(x^5)=x^{10}+x^5+1$

$f_2(x^5)$, when divided by $f_2(x)$ leaves a remainder 0.

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