59th Ukrainian National Mathematical Olympiad

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: $f(x)(x-t)=g(x)h(x)$. $$(x^3-x^2+cx+4)(x-t)=(x^2-3x+a)(x^2+x+2).$$ 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: $$f(x)=x^3-x^2-4x+4,g(x)=x^2-3x+2,h(x)=x^2+x-2.$$