Why are the coefficients always equal? Take the equation $ax^{2} + bx + c = 3x^{2} + 4x + 53$.
Why is it always true that $a = 3, b = 4$ and $c = 53$?
I've seen many examples like this where the coefficients are equated, and was just wondering why that is always true.
 A: The equation is valid for all $x$ take $x=0$ to get $c=53$. The equation now rewrites as follows $ax^2+bx=3x^2+4x$ and again this is true for all $x$ especially for say $x=-{4\over 3}$ and the equation now looks like ${16a\over 9}-{4b\over 3}=0$ and for $x=1$ we get $a+b=7$ the solution of the system in $(a,b)$ is $(3,4)$
In general if two polynomial are such as $a_nx^n+\cdots+a_0=b_nx^n+\cdots+b_0$ for all x this means $(a_n-b_n)x^n+\cdots+(a_0-b_0)$ has an infinity of solutions while it has degree $n$ and therefore all its coefficients are $0$ which translates into $a_i=b_i\,\forall i$
A: Here's a cute way of proving it I don't think has been mentioned yet.
If you agree two polynomials are equal, you should also agree that their derivatives are equal. 
Let $f(x) = ax^2 + bx + c = 3x^2+4x+53$.
Taking the derivative of $f$, we get:
$f'(x) = 2ax + b = 6x + 4$.
Taking the second derivative, we get:
$f''(x) = 2a = 6$.
Working backwards, we can solve for each variable in turn. The last equation gives $a = 3$. Going to the one before it, we can replace $2ax$ with $6x$ and subtract $6x$ from both sides to get $b=4$. And finally, we go back to the original to subtract $3x^2 + 4x$ from both sides to get $c=53$.
It's not the most efficient way, but it doesn't require any evaluation of the polynomial or talk of its roots.
A: The set $\{1,x,x^2,...\}$ is a linear independent set in the vector space of real valued polynomials ($+$ and $\cdot$ defined in the obvious way). Then, by definition of linear independence,
$$
\sum_{i=0}^n a_ix^i = \sum_{j=0}^m b_jx^j \iff a_i = b_j \quad \forall i,j<n=m.
$$
A: Suppose that $ax^{2} + bx + c = 3x^{2} + 4x + 53$ for all $x$, or $ax^{2} + bx + c - (3x^{2} + 4x + 53) = 0$ for all $x$. 
$ax^{2} + bx + c - (3x^{2} + 4x + 53) = 0$ is a polynomial with at most degree $2$ so by the fundamental theorem of algebra it has at most $2$ roots if it is not the zero polynomial. 
But $ax^{2} + bx + c - (3x^{2} + 4x + 53) = 0$ has infintely many roots (it is zero for every value of $x$), so it must be the zero polynomial. Then we have $a=3$, $b=4$ and $c=53$. 
$$ax^{2} + bx + c - (3x^{2} + 4x + 53) = (a-3)x^2+(b-4)x+(c-53)$$
Every coefficient must be equal to zero for it to be the zero polynomial, thus $a-3=0$, and $a=3$. Similiarly, one has $b=4$ and $c=53$. 
A: To try and give a more general answer to why $a=d$, $b=e$ and $c=f$ given that $ax^{2} + bx + c = dx^{2} + ex + f$. 
Rearranging gives $ax^{2} + bx + c - (dx^{2} + ex + f) = 0$, combining factors of $x$ gives $(a-d)x^{2} + (b-e)x + (c-f) = 0$.
It then follows that each of $(a-d)x^{2}$, $(b-e)x$ and $(c-f)$ must equal $0$.
If $x$ can have any value then $a - d = 0$ therefore $a = d$ with the same for $b = e$ and $c = f$.
It should be noted that values can be found that mean $a$ and $d$, $b$ and $e$, and $c$ and $f$ are not equal for specific values of $x$. Such as $x$ = $0$, which allows $a$, $b$, $d$ and $e$ to have any value.
