1
$\begingroup$

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 $−2$?

$\endgroup$
  • $\begingroup$ We have $ax^3+x^2(b-3a)+x(2a-3b)+2b$. Then I will show by contradiction that at least one of its coefficients is equal to or less than $-2$. Assume on the contrary the opposite. Then $b \geq -1, a \geq -1$. Then what? $\endgroup$ – user19405892 Dec 6 '15 at 15:55
  • $\begingroup$ We have $ax^3+x^2(b-3a)+x(2a-3b)+2b$. If $a$ is negative we have $ax^3+x^2(b+3a)+x(-2a-3b)+2b$ for $a \geq 1$. We then see that $a$ can't be negative otherwise $x(2a-3b)$ will have a coefficient that is less than or equal to $-2$. Therefore, $a \geq 1$. We also have that $2b > -2$ and $b \geq 1$. $\endgroup$ – user19405892 Dec 6 '15 at 16:08
  • $\begingroup$ @turkeyhundt $a$ is an integer. $\endgroup$ – rogerl Dec 6 '15 at 16:09
  • $\begingroup$ We have $b-3a > -2; 2a-3b > -2$ and from that we get $ 1 \leq a < \dfrac{8}{7}$ implying that $a = 1$. Our polynomial then becomes $x^3+x^2(b-3)+x(2-3b)+2b$ and from which we obtain $1 < b < 4/3$, a contradiction. Therefore, at least one of its four coefficients is equal to or less than $-2$. $\endgroup$ – user19405892 Dec 6 '15 at 16:15
  • $\begingroup$ @rogerl Ha. It pays to read the original question. Thx. $\endgroup$ – turkeyhundt Dec 6 '15 at 16:20
0
$\begingroup$

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 $\le -2$

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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.)

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.