Given that (x-3) is a factor of F(x) = ax3 + bx2 − 9x − 9, we must have F(3) = 0.

The Factor Theorem states that if $(x-c)$ is a factor of a polynomial $F(x)$, then $F(c) = 0$.
Given that $(x-3)$ is a factor of $F(x) = ax^3 + bx^2 - 9x - 9$, we must have $F(3) = 0$.

Substitute $x=3$ into the polynomial $F(x)$:

Since $F(3) = 0$:

Divide the entire equation by 9:

Rearrange the equation to match the required form:

This shows that $3a + b = 4$.

b) When $F(x)$ is divided by $(x-2)$, the remainder is $-15$. This is a given condition. By the Remainder Theorem, this means $F(2) = -15$.