Yes and no. A cubic Bezier curve is, in the end, a cubic polynomial. As is a cubic Hermite curve, etc. The difference is: what are the degrees of freedom that control this polynomial? In the slide above, you have four coefficients $a,b,c,d$ that control four homogeneous terms $t^3$, $t^2$, $t$, and $1$, respectively. With a cubic Bezier curve you have four control points $p_0, p_1, p_2, p_3$ that control four inhomogeneous Bernstein bases. For any given set of coefficients $a,b,c,d$ you can find a corresponding $p_0,p_1,p_2,p_3$ and vice versa (by a linear transformation). So, you can (and should) think of different curve representations as different coordinate systems for the space of cubic polynomials.
Isn't this the bezier curve?
Yes and no. A cubic Bezier curve is, in the end, a cubic polynomial. As is a cubic Hermite curve, etc. The difference is: what are the degrees of freedom that control this polynomial? In the slide above, you have four coefficients $a,b,c,d$ that control four homogeneous terms $t^3$, $t^2$, $t$, and $1$, respectively. With a cubic Bezier curve you have four control points $p_0, p_1, p_2, p_3$ that control four inhomogeneous Bernstein bases. For any given set of coefficients $a,b,c,d$ you can find a corresponding $p_0,p_1,p_2,p_3$ and vice versa (by a linear transformation). So, you can (and should) think of different curve representations as different coordinate systems for the space of cubic polynomials.