Divide $2 t^{4}+3 t^{3}-2 t^{2}-9 t-12$ by $t^{2}-3$.

(A) The dividend is $p(t) = 2t^4 + 3t^3 - 2t^2 - 9t - 12$ and the divisor is $s(t) = t^2 - 3$.
Step $1$: Divide the first term of the dividend $(2t^4)$ by the first term of the divisor $(t^2)$ to get $2t^2$. Multiply $2t^2$ by $(t^2 - 3)$ to get $2t^4 - 6t^2$. Subtract this from the dividend.
Step $2$: The result is $3t^3 + 4t^2 - 9t - 12$. Divide $3t^3$ by $t^2$ to get $3t$. Multiply $3t$ by $(t^2 - 3)$ to get $3t^3 - 9t$. Subtract this from the current expression.
Step $3$: The result is $4t^2 - 12$. Divide $4t^2$ by $t^2$ to get $4$. Multiply $4$ by $(t^2 - 3)$ to get $4t^2 - 12$. Subtract this to get a remainder of $0$.
Thus,the quotient is $2t^2 + 3t + 4$ and the remainder is $0$.