30.2. Given that $x+2$ and $x-3$ are factors of $p(x)=a x^{3}+a x^{2}+b x+12$, what is the remainder when $p(x)$ is divided by $x-1$ ?
30.2. 18
Since $x+2$ is a factor of $p(x)=a x^{3}+a x^{2}+b x+12$, Factor Theorem guarantees that
$$
p(-2)=-8 a+4 a-2 b+12=0 \quad \text { or } \quad 2 a+b=6 .
$$

Similarly, we also have
$$
p(3)=27 a+9 a+3 b+12=0 \quad \text { or } \quad 12 a+b=-4 .
$$

Solving the system of equations involving $(1 \star)$ and $(2 \star)$, we get $a=-1$ and $b=8$. Thus, we have $p(x)=-x^{3}-x^{2}+8 x+12$, so that $p(1)=18$.