So given $ax^2 + bx + c = dx^2 + ex + f$ and that it holds true for all values of x:
Prove $a = d$, $b = e$, and $c = f$.
What I have done so far is set the equation equal to zero and factor the desired variables together and out:
$ax^2 + bx + c - dx^2 - ex - f = 0$
$ax^2 - dx^2 + bx - ex + c - f = 0$
$(a-d)x^2 + (b-e)x + (c-f) = 0$
Now once I prove that $a - d$, $b - e$, and $c - f$ are all equal to zero, then everything would be all over.
But unfortunately I do not know how to proceed here. I know it probably has to do with polynomial degree root limits or something, but I am fuzzy on the precise procedure and would appreciate the help on finishing this off.
.Since the equation: $$ (a-d)x^2 + (b-e)x+(c-f)=0 $$ is true for all values of $x$, we can substitute values of $x$ and the resulting equation will still be true:
Put $x=0$, then we get $c-f=0$, so $c=f$.
Now our equation simplifies to : $$ (a-d)x^2 + (b-e)x=0 \implies x((a-d)x+(b-e)) = 0 $$
Take $x=1$ here: $(a-d) + (b-e) = 0$
Take $x=-1$ here:$(d-a) + (b-e) = 0$
Add the two equations: $2(b-e) = 0$, so $b-e=0$ and $b=e$
Finally, we are left with $(a-d)x^2=0$ for all values of $x$. Put $x=1$ to get $a-d=0$ so $a=d$. This method required no kind of complicated method and can be generalized to other polynomials.