Prove that if $lx^2+mx+n$ is equal to zero for three distinct values then $l=m=n=0$

Question

How should I go around to prove that if $lx^2+mx+n$ equals to zero for three distinct values of $x$, then $l=m=n=0$?

I tried letting $P(x)=lx^2+mx+n$ and used the factor theorem, but I can't seem to reveal the result.

Answer 1

This polynomial has degree 2, therefore it can not have 3 distinct roots, hence all the coefficients must be zero

Answer 2

Assume there are 3 distinct roots $r_1,r_2,r_3$. Then we have the system of equations $$lr_1^2+mr_1+n=0\\lr_2^2+mr_2+n=0\\lr_2^2+mr_2+n=0\\$$ as $r_1\neq r_2\neq r_3$ we can add and subtract lines to get $$l(r_1+r_2)+m=0\\l(r_2+r_3)+m=0\\lr_2^2+mr_2+n=0$$ but then $l(r_1+r_2)=l(r_2+r_3)$ which is onlye possible if $l=0$. If $l=0$ then $mr_1=mr_2$ which again yields that $m=0$. And so we also have $n=0$.