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.
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A polynomial of degree $d$ with coefficients in a field has at most $d$ roots. – Bernard Jun 27 '21 at 09:42
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Hint: Fundamental theorem of algebra. – lonza leggiera Jun 27 '21 at 09:44
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1More explicitly, if $f(x)$ is a polynomial of degree $n$, with root $r$ [i.e. where $f(r) = 0$], then there exists a polynomial $g(x)$ of degree $(n-1)$ such that $f(x) = (x-r)g(x).$ See, for example, this answer. – user2661923 Jun 27 '21 at 10:54
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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 $n=0$.
cansomeonehelpmeout
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This polynomial has degree 2, therefore it can not have 3 distinct roots, hence all the coefficients must be zero
Huy Nguyen
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