Given P(x) = (2x-1)(x-3)Q(x) + 12x - 8, where P(x) is a polynomial of degree 3. If P(x) is exactly divisible by x-1 and P(0)=10, find Q(x).

The given polynomial is $P(x) = (2x-1)(x-3)Q(x) + 12x - 8$.
The term $(2x-1)(x-3)$ is a quadratic, which has a degree of 2.
We are given that $P(x)$ is of degree 3.
For the product $(2x-1)(x-3)Q(x)$ to result in a polynomial of degree 3 (after considering the highest degree term), $Q(x)$ must be a linear polynomial (degree 1).
Let $Q(x) = ax+b$.

By the Factor Theorem, if $P(x)$ is exactly divisible by $x-1$, then $P(1) = 0$.
Substitute $x=1$ into the expression for $P(x)$:

Since $P(1) = 0$:

Substitute $x=0$ into the expression for $P(x)$:

Since $P(0) = 10$:

We have $Q(x) = ax+b$.
Using $Q(0) = 6$:

So, $Q(x) = ax+6$.

Using $Q(1) = 2$:

Therefore, $Q(x) = -4x+6$.

The final answer is $-4x+6$.