If the remainder on division of $x^{3}+2 x^{2}+k x+3$ by $x-3$ is $21$,find the quotient and the value of $k$. Hence,find the zeroes of the cubic polynomial $x^{3}+2 x^{2}+k x-18$.

(A) Let $p(x) = x^{3}+2 x^{2}+k x+3$.
By the Remainder Theorem,since $p(x)$ is divided by $x-3$,the remainder is $p(3)$.
Given $p(3) = 21$,we have:
$3^{3}+2(3)^{2}+k(3)+3 = 21$
$27 + 18 + 3k + 3 = 21$
$48 + 3k = 21$
$3k = 21 - 48 = -27$
$k = -9$.
Now,dividing $x^{3}+2 x^{2}-9 x+3$ by $x-3$ using long division:
$x^{3}+2 x^{2}-9 x+3 = (x-3)(x^{2}+5x+6) + 21$.
The quotient is $x^{2}+5x+6$.
To find the zeroes of $x^{3}+2 x^{2}-9 x-18$,we factorize it:
$x^{3}+2 x^{2}-9 x-18 = x^{2}(x+2) - 9(x+2)$
$= (x^{2}-9)(x+2)$
$= (x-3)(x+3)(x+2)$.
Setting the polynomial to $0$,the zeroes are $x = 3, -3, -2$.