1. ## Harder Polynomial Question

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.

2. Originally Posted by xwrathbringerx
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.

Have you tried subbing in the values of 1 and 3, see what you get?

3. Ummm

P(1) = 1^p.3^q - 1^p + 1^q -1^q.3^p+3^p-3^q
P(3)=0

I've got these but I have no idea how to use that to demonstrate what is required ...

Is it because P(3) = 0, it means (x-3) is a factor and SO (x-1)(x-3) is???

4. Hello, xwrathbringerx!

You had it . . . and didn't know it.

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.
Fact: .If $P(a) = 0$, then $(x-a)$ is a factor of $P(x).$

$P(1) \:=\:1^p(3^q-1) + 1^q(1-3^p) + 3^p - 3^q \;=\; 3^q - 1 + 1 - 3^p + 3^p - 3^q \;=\;0$

. . Therefore, $(x-1)$ is a factor of $P(x).$

$P(3) \;=\;3^p(3^q - 1) + 3^q(1-3^p) + 3^p - 3^q \;=\;3^{p+q} - 3^p + 3^q - 3^{p+q} + 3^p - 3^q \;=\; 0$

. . Therefore, $(x-3)$ is a factor of $P(x).$