(A) To check if $t^{2}-3$ is a factor of $2t^{4}+3t^{3}-2t^{2}-9t-12$,we perform polynomial long division.
We write the divisor as $t^{2}+0t-3$.
Dividing $2t^{4}+3t^{3}-2t^{2}-9t-12$ by $t^{2}+0t-3$:
$1$. Divide $2t^{4}$ by $t^{2}$ to get $2t^{2}$. Multiply $2t^{2}(t^{2}+0t-3) = 2t^{4}+0t^{3}-6t^{2}$. Subtracting this from the dividend gives $3t^{3}+4t^{2}-9t-12$.
$2$. Divide $3t^{3}$ by $t^{2}$ to get $3t$. Multiply $3t(t^{2}+0t-3) = 3t^{3}+0t^{2}-9t$. Subtracting this gives $4t^{2}+0t-12$.
$3$. Divide $4t^{2}$ by $t^{2}$ to get $4$. Multiply $4(t^{2}+0t-3) = 4t^{2}+0t-12$. Subtracting this gives a remainder of $0$.
Since the remainder is $0$,$t^{2}-3$ is a factor of $2t^{4}+3t^{3}-2t^{2}-9t-12$.