Answer
The equality $ax^2+bx+c=0$ is satisfied for all $x$ when $a=b=c=0$. We have to show that there can be no other choice of $a, b, c$ and we will do that by showing that for three particular values of $x$ this choice is unique. Check the step by step guide for details.
Work Step by Step
+Let us make a system of equation by choosing three different $x$-es and and show that this system has a unique solution $a=b=c=0$:
For $x=0$ we have $a\times 0^2 + b\times 0 + c=0\Rightarrow c=0$
For $x=1$ we have $a\times 1^2 + b\times 1 + c =0\Rightarrow a+b+c=0$
For $x=-1$ we have $a(-1)^2 + b(-1)+c=0\Rightarrow a-b+c=0$
So now we have the following system:
\begin{align*}
a+b+c=&0\\
a-b+c=&0\\
c=&0
\end{align*}
The last equation already gives the solution for $c$:
$$c=0$$
Putting this into the firs two we get
\begin{align*}
a+b=&0\\
a-b=&0.
\end{align*}
Adding these two together we obtain
$$a+a+b-b=0\Rightarrow 2a=0$$ and this gives a solution for $a$:
$$a=0$$
Putting this into the first equation we have
$$0+b=0\Rightarrow b=0.$$
Now to return to the original problem, the equality $ax^2+bx+c=0$ is satisfied for all $x$ when $a=b=c=0$. We have shown that there can be no other choice of $a, b, c$ by showing that for three particular values of $x$ this choice is unique.
