Prove that the cubic has at least one of its four coefficients equals to or less than $-2$. Suppose that $P(x)$ is a polynomial of degree $3$ with integer coefficients and that $P(1)=0,P(2)=0$. Prove that at least one of its four coefficients is equal to or less than $−2$.
We can write such a polynomial as $(x-1)(x-2)(ax+b) = a x^3-3 a x^2+2 a x+b x^2-3 b x+2 b$. What is the next step to prove that at least one of its four coefficients is equal to or less than $−2$?
 A: Let $P(x)=-(ax^3+bx^2+cx+d)$, where the negative sign out front allows us to reformulate the condition as one of the integer coefficients $a$, $b$, $c$, or $d$ is at least $2$.  Now suppose the condition is not met, so $a,b,c,d\le1$.
The assumption $P(1)=0$ implies $a+b+c+d=0$.  There are four possibilities:
$$\{a,b,c,d\}=
\begin{cases}
\{1,1,1,-3\}\\
\{1,1,-1,-1\}\\
\{1,0,0,-1\}\\
\{1,1,0,-2\}\\
\end{cases}$$
The first two cases can be dismissed because if all four coefficients are odd, then $P(2)$ is odd.  The third case can be dismissed because no two powers of $2$ are equal, and the fourth case can be dismissed because no two powers of $2$ sum to another power of $2$.
Note, this proof applies, with a tiny extra argument, to any polynomial with at most $4$ nonzero (integer) coefficients.  (The extra argument is to factor out and ignore the largest power of $x$ dividing the polynomial.)
A: Necessarily there is a third real root $x_0$. Then one has $$f(x)=x^3-(3-x_0)x^2+(2+3x_0)x-2x_0  $$ We can verify that the system of inequalities
$$-3+x_0>-2\iff x_0>1$$ and $$-2x_0>-2\iff x_0<1$$ is incompatible so one of the two considered coefficients must be $\le -2$
