4. The quadratic trinomial $a x^{2}+b x+c$ has two distinct roots, and its coefficients are non-zero integers. Moreover, the product of all its coefficients equals the product of all its roots. What values can this product take? Provide a complete and justified answer to this question.
4. Answer: The product can be any negative integer. Solution From the condition of the problem and Vieta's theorem, it follows that $\mathrm{a}^{2} \mathrm{bc}=\mathrm{c}$. Since all coefficients are integers and non-zero, this means that $\mathrm{b}=1, \mathrm{a}= \pm 1$ and the quadratic trinomial has the form either $x^{2}+x+c$, or $-x^{2}+x+c$. In the first case, the product of the roots is $\mathrm{C}$, and in the second case, it is $-C$. The presence of two distinct roots in the first case is equivalent to the inequality $c<1/4$, and in the second case, to the inequality $-1 / 4 \leq c$. In each of these cases, the same result is obtained: the desired product can be any negative integer.

Criteria: Full solution with explanation - 7 points. Both constructions are provided, the condition is verified, and the correct answer is given - 3 points. The condition of the positivity of the discriminant is not taken into account - no more than 2 points. The case of the second trinomial is not considered - 2 points are deducted.