Polynomial $P(x)$ contains only terms of odd degree.When $P(x)$ is divided by $(x-3)$ ,the remainder is $6$. Polynomial $P(x)$ contains only terms of odd degree.When $P(x)$ is divided by $(x-3)$ ,the remainder is $6$.If $P(x)$ is divided by $(x^2-9)$ then remainder is $g(x).$Find the value of $g(2).$

As $P(x)$ is a polynomial containing only terms of odd degree.Therefore it should pass through origin.
$P(0)=0$
When $P(x)$ is divided by $(x-3)$ ,the remainder is $6$.
$P(x)=Q_1(x)(x-3)+6....................(1)$
If $P(x)$ is divided by $(x^2-9)$ then remainder is $g(x)$.
$P(x)=Q_2(x)(x^2-9)+ax+b............(2)$,where $ax+b=g(x)$
From $(1),P(3)=6$
From $(2),P(3)=6=3a+b$
But we need to find $g(2)=2a+b$
I dont know how to solve further.Please help me.Thanks.

Since it is an odd degree polynomial
P(–a)= –p(a)
Let P(x)=(x^–9)Q(x)+ax+b
(where Q is quotient and (ax+b) = g(x) = remainder)
Given P(3)=6
Therefore P(-3) = -6
Solve these two equations to get value of a and b
We get a=2 and b=0
G(x) =ax+b
G(2)= 2(2)+0 =4
