Question

The polynomial $R ( x )$ is the remainder upon dividing $x ^{2007} byx ^2-5 x +6$. If $R (0)$ can be expressed as $a b\left(a^c-b^c\right)$, find the value of $(a+b+c)$.

Answer 1

$\text { now } R (0)= b =? $
$\text { Put } x =2 \text { in }(1)$
$ 2 a + b =2^{2007}$....(2)
$\text { put } x =3 \text { in }(1) $
$ 3 a + b =3^{2007}$....(3)
(3)-(2) gives
$a=3^{2007}-2^{2007}$
now $b \left.=2^{2007}-2 a=2^{2007}-2\left(3^{2007}-2^{2007}\right)=2^{2007}+2 \cdot 2^{2007}-2 \cdot 3^{2007}\right) $
$ =3 \cdot 2^{2007}-2 \cdot 3^{2007}=6\left[2^{2006}-3^{2006}\right]=2 \cdot 3\left[2^{2006}-3^{2006}\right] \equiv a b\left(a^c-b^c\right)$
$\text { hence } a=2 ; b=3 ; c=2006 $
$\therefore a+b+c=2+3+2006=2011$