To begin with, you are dealing with the "great old theory" of **elimination**.

In this case, the answer is : yes, it is always possible to eliminate parameter $t$ between polynomial expressions and obtain an implicit polynomial expression. 

Let us show it on an example, with a parametric curve defined by these equations :

$$\tag{1}\begin{cases}x=t^2+t+1\\y=t^3-1\end{cases}$$

If you use a Computer Algebra System such as Mathematica, elimination process is implemented under the following form: Eliminate[{x==t^2+t+1,y==t^3-1},t] giving the following implicit equation : 

$$\tag{2}x^3-3x^2-3xy-y^2=0$$

as its result.

But there is more to say. You need for this to be familiar with **resultants** (see remark below) (https://en.wikipedia.org/wiki/Resultant). The theorem you are looking for is the fact that the implicit equation can be obtained as the resultant of the $\color{red}{2}$nd degree polynomial $t^2+t+(1-x)$ (I was tempted to write $=0$...) and $\color{red}{3}$rd degree polynomial $t^3-(1+y)$ which the $5 \times 5$ parametric determinant (5 = $\color{red}{2+3}$):

$$d=\left|\begin{array}{ccccc}1& 1& (1-x)& 0 &0 \\
 0& 1& 1& (1-x)& 0\\
 0& 0& 1& 1& (1-x)\\
 1& 0& 0& -(1+y)& 0\\
 0& 1& 0& 0& -(1+y)\end{array}\right|=0$$

(one writes $\color{red}{3}$ times the $\color{red}{2}$nd degree polynomial, then $\color{red}{2}$ times the $\color{red}{3}$rd degree one, with a right shift for every "carriage return").

The expansion of determinant $d$ gives back (up to a certain unimportant constant factor) implicit equation (2).

This example can evidently been extended to a general case. 

**Remark:** the main idea behind the concept of resultant is that it expresses a necessary and sufficient condition between parameter(s) present in coefficients of 2 equations for these equations to have at least a common root. Let us take an example : consider a quadratic polynomial $at^2+bt+c$ (with $a\ne0$) and its derivative $2at+b$. They have a common root $t_0$ iff the following resultant is zero:


 $$\delta=\left|\begin{array}{ccc}a& b& c \\
 2a& b& 0\\
 0& 2a& b\end{array}\right|=-a(b^2-4ac)$$

This value is $0$ iff $b^2-4ac=0$, which is the classical criteria for a quadratic equation to have a double root. We have thus found back the concept of discriminant by using the resultant!