Since $f(x)$ is a cubic polynomial, it has no more than three real roots, hence, quadratic polynomials $g(x)$ and $h(x)$ have the same root. Let us denote this root by $t$. Then,
$$g(t) = t^2 - 3t + a = 0 \text{ and } h(t) = t^2 + t + b = 0 \Rightarrow 4t + b - a = 0 \Rightarrow t = \frac{1}{4}(a-b).$$

Then, the equality must hold: $f(x)(x-t) = g(x)h(x)$.
$$(x^3 - x^2 + c x + 4)(x - t) = (x^2 - 3x + a)(x^2 + x + 2).$$
By collecting coefficients of $x^3$, we obtain that the equation must be satisfied:
$$-1-t = -3+1 \Rightarrow t=1, \text{ hence, } f(1) = f(t) = 0.$$

It is easy to find an explicit form of polynomials $f$, $g$ and $h$, which satisfy the given statement:
$$f(x) = x^3 - x^2 - 4x + 4, \quad g(x) = x^2 - 3x + 2, \quad h(x) = x^2 + x - 2.$$