What is equating the coefficient of the corresponding power of x? I recently asked a question on mathematics.SE pertaining to solving an unusual quadratic [1], and was introduced to the phrase 'equating the coefficient of the corresponding power of x'. What does this mean?
 A: Simply put, if you have an equation where each side is a polynomial, like so:
$$A_nx^n + A_{n-1}x^{n-1}+...+A_1x+A_0 = B_nx^n + B_{n-1}x^{n-1}+...+B_1x+B_0$$
then "equating coefficients" means recognizing that the equation holds if and only if $A_i = B_i$ for $i = 0,1,...,n$.
An easy way to see why this is true is to use derivatives. For $i = 0,1,...,n$, let $C_i = A_i - B_i$. Then $A_i = B_i$ if and only if $C_i = 0$. From the equation above we have
$$C_nx^n+C_{n-1}x^{n-1}+...+C_1x+C_0 = 0$$
Taking derivatives of both sides yields
$$nC_nx^{n-1}+(n-1)C_{n-1}x^{n-2}+...+2C_2x+C_1 = 0$$
Taking the second derivative gives
$$n(n-1)C_nx^{n-2}+(n-1)(n-1)C_{n-1}x^{n-3}+...+(3)(2)C_3x+2C_2 = 0$$
Continuing on to the $n$-th derivative gives
$$n(n-1)\cdots(2)(1)C_n = n!C_n = 0$$
from this we see clearly that $C_n = 0$. Now we simply go backwards through the derivatives.
The $(n-1)$-st derivative was 
$$n(n-1)\cdots(2)C_nx+(n-1)!C_{n-1} = 0$$
Since we know that $C_n = 0$, we can simplify this to
$$(n-1)!C_{n-1} = 0$$
thereby concluding that $C_{n-1}=0$ as well.
We can continue going back through all the derivatives, concluding at each step that the corresponding remaining coefficient must be zero. By the time we've arrived back at the original equation, we've shown that $C_n=C_{n-1}=...=C_1=C_0=0$.
A: It means that if you know that $$ax^2+bx+c = dx^2 +ex + f$$ holds for all $x$, so that the two functions are identical, then one must have $a=d, b=e, c=f$—one can equate the coefficients of the two polynomials.
In particular, if one knows $$px^2 + qx + r=0$$ holds for all $x$, then one must have $p=q=r=0$.
(You might like to prove that this is actually true.)
A: From looking at the linked question user15... is referring to the coefficients of a particular power of $x$ on one side of the equality are equal to the corresponding coefficients of the same power of $x$ on the other side of the equality, i.e. if I have $3x^2+2x-1=ax^2+bx+c$, then $a=3, 2=b, c=-1$ since those are the corresponding powers of $x$. 
