I'm interesting in trying to solve the intersection points for a cubic Bézier curve with a line segment.
Background
A point on a cubic Bézier curve is given by,
$$ P_b(t_b) = \left[ \begin{matrix} 1,t_b, t_b^2, t_b^3 \end{matrix} \right] M_b \left[ \begin{matrix} p_0, p_1, p_2, p_3 \end{matrix} \right]^T $$
where $M_b$ is a $4\times 4$ matrix containing the Bernstein polynomial coefficients for order $n=3$.
Similarly, the equation for a point on the line segment is given by,
$$ P_l(t_l) = \left[ \begin{matrix} 1,t_l \end{matrix} \right] M_l \left[ \begin{matrix} p_4, p_5\end{matrix} \right]^T $$
where $B_l = \left[ \begin{matrix} 1, 0 \\ -1, 1\end{matrix} \right]$ is is a $2\times 2$ matrix containing coefficient for $n=1$.
Here $t_b$ and $t_l$ are dimensionless distance (i.e. a number between 0 and 1) along the Bézier and the line-segment respectively; $p_i = (x_i, y_i)$ represents a point in a 2D space.
At the intersection points,
$$ P_b(t_b) - P_l(t_l) = 0 $$
The solution should emit pairs of $t_b$ and $t_l$ values when the curve and lines cross.
Attempt
Expanding the matrix expressions and explicitly writing the $x$ and $y$ components leads to a simultaneous equation of form,
$\text{cubic}_x(t_b) - \text{linear}_x(t_l) = 0 \\ \text{cubic}_y(t_b) - \text{linear}_y(t_l) = 0 $
which should be solved to find pairs of $t_b$ and $t_l$.
Q In general should I expect that a closed form expression is possible for such a system?
The actual equations are,
$$ \left(p_{0,x}(-x^3 + 3x^2 - 3x + 1) + p_{1,x}(3x^3 - 6x^2 - 3x) + p_{2,x}(-3x^3 + 3x^2) + p_{3,x}x^3\right) - \left(p_{4,x}(-y + 1) + p_{5,x}y\right) = 0 $$
$$ \left(p_{0,y}(-x^3 + 3x^2 - 3x + 1) + p_{1,y}(3x^3 - 6x^2 - 3x) + p_{2,y}(-3x^3 + 3x^2) + p_{3,x}x^3\right) - \left(p_{4,x}y + p_{5,x}(-y + 1)\right) = 0 $$
where $p_0 = (p_{0,x}, p_{0,y})$
Q Does this specific set of equations have an analytical, closed form, solution?
It's much easier if you write the equation of the line in implicit form. You can eliminate the parameter $t_l$ from your line equation, and get an equation of the form $ax+by+c=0$.
Substitute the $x$ and $y$ components of your Bezier curve equation into the line equation, and you'll get a single cubic equation for $t_b$. Then, there is a formula that gives you the solutions of this cubic (though rather an ugly one).
Once you have found an intersection point, you need to check that it lies on the line segment (between the points $p_4$ and $p_5$).
For more on how to intersect curves, take a look at this site. The technique I described is case 5.6.1.1.
In general, the theory of elimination and resultants tells us whether or not sets of algebraic equations have solutions. And, in fact, resultants are often used to do curve intersection calculations (in academia, anyway).