Show that (x-1)(x-3) is a factor of P(x) = x^p(3^q-1)+x^q(1-3^p)+(3^p-3^q) where p and q are positive integers.
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Fact: .If $\displaystyle P(a) = 0$, then $\displaystyle (x-a)$ is a factor of $\displaystyle P(x).$Show that $\displaystyle (x-1)(x-3)$ is a factor of: .$\displaystyle P(x) =\: \:x^p(3^q-1) + x^q(1-3^p) + 3^p-3^q$
where $\displaystyle p$ and $\displaystyle q$ are positive integers.
$\displaystyle P(1) \:=\:1^p(3^q-1) + 1^q(1-3^p) + 3^p - 3^q \;=\; 3^q - 1 + 1 - 3^p + 3^p - 3^q \;=\;0 $
. . Therefore, $\displaystyle (x-1)$ is a factor of $\displaystyle P(x).$
$\displaystyle P(3) \;=\;3^p(3^q - 1) + 3^q(1-3^p) + 3^p - 3^q \;=\;3^{p+q} - 3^p + 3^q - 3^{p+q} + 3^p - 3^q \;=\; 0$
. . Therefore, $\displaystyle (x-3)$ is a factor of $\displaystyle P(x).$