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Find the remainder when $x^{51} + 51$ is divided by $x-a$. The given answer is 50 but am not reaching it.

Please help, with suitable explanation.

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  • $\begingroup$ This is a new topic for me so don't know watta do. $\endgroup$ – user4845219 Apr 29 '15 at 12:25
  • $\begingroup$ Look up the Remainder Theorem. $\endgroup$ – tomi Apr 29 '15 at 12:28
  • $\begingroup$ How to implement that theorem here ?? $\endgroup$ – user4845219 Apr 29 '15 at 12:30
  • $\begingroup$ note that from $P(x)=Q(x)(x-a)+t$ hence $t=P(a)$ $\endgroup$ – Elaqqad Apr 29 '15 at 13:18
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The remainder factor theorem states the following:

If we have a polynomial $P(x)$, then the remainder we get on dividing $P(x)$ by $(ax-b)~,~a\neq 0$ is given by $P\left(\frac{b}{a}\right)$ That is,

$$P(x)\equiv P\left(\frac{b}{a}\right)\pmod{ax-b}~,~a\neq 0$$

Proof:

Take $Q(x)$ as the quotient polynomial on division of $P(x)$ by $ax-b$. Then,

$$P(x)=(ax-b)Q(x)+\textrm{Rem}$$

Plug in $x=\dfrac{b}{a}$ here to get,

$$P\left(\frac{b}{a}\right)=\textrm{Rem}$$


Using this theorem, you will get that the remainder is $a^{51}+51$ which is dependent on the value of $a$.

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