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

The polynomial is given by $P(x) = (2x-1)(x-3)Q(x) + 12x-8$.

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

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

The polynomial $P(x)$ has a degree of 3.
The term $(2x-1)(x-3)$ is a quadratic expression (degree 2).
The term $12x-8$ is a linear expression (degree 1).
For $P(x)$ to be of degree 3, the product $(2x-1)(x-3)Q(x)$ must be of degree 3.
Since $(2x-1)(x-3)$ is degree 2, $Q(x)$ must be a linear polynomial (degree 1).
Let $Q(x) = ax+b$.

From $Q(1)=2$:

From $Q(0)=6$:

Substitute $b=6$ into equation $(*)$:

Using $a=-4$ and $b=6$: