For which values of $a$ and $b$ are the zeroes of $q(x) = x^{3} + 2x^{2} + a$ also the zeroes of the polynomial $p(x) = x^{5} - x^{4} - 4x^{3} + 3x^{2} + 3x + b$? Which zeroes of $p(x)$ are not the zeroes of $q(x)$?

(A) Given that the zeroes of $q(x) = x^{3} + 2x^{2} + a$ are also the zeroes of $p(x) = x^{5} - x^{4} - 4x^{3} + 3x^{2} + 3x + b$,it implies that $q(x)$ is a factor of $p(x)$.
Performing polynomial long division of $p(x)$ by $q(x)$:
Dividing $x^{5} - x^{4} - 4x^{3} + 3x^{2} + 3x + b$ by $x^{3} + 2x^{2} + a$ yields a quotient of $x^{2} - 3x + 2$ and a remainder of $-(1+a)x^{2} + (3+3a)x + (b-2a)$.
Since $q(x)$ is a factor,the remainder must be zero for all $x$,so the coefficients must be zero:
$-(1+a) = 0 \Rightarrow a = -1$
$3+3a = 0 \Rightarrow 3+3(-1) = 0$ (Consistent)
$b-2a = 0 \Rightarrow b = 2(-1) = -2$.
Thus,$a = -1$ and $b = -2$.
Substituting these values,$p(x) = (x^{3} + 2x^{2} - 1)(x^{2} - 3x + 2)$.
Factoring the quotient: $x^{2} - 3x + 2 = (x-1)(x-2)$.
The zeroes of $p(x)$ are the zeroes of $q(x)$ along with the zeroes of the quotient,which are $x=1$ and $x=2$. Therefore,the zeroes of $p(x)$ that are not zeroes of $q(x)$ are $1$ and $2$.