Equality of polynomials We know that $a_nx^n + a_{n-1}x^{n-1} + \dots + a_0 = b_nx^n + b_{n-1}x^{n-1} + \dots + b_0 $. How we can prove that $a_n = b_n , a_{n-1} = b_{n-1} , \dots ,a_0 = b_0$ . Also if in right side instead of $x$ we put $z$ is this statement true ? $a_n = b_n , a_{n-1} = b_{n-1} , \dots ,a_0 = b_0$ 

That depends.


*

*If you consider them as formal polynomials, then the equality of coefficients is by definition.

*If you consider them as functions $\mathbb R\to\mathbb R$, you can insert $n+1$ different values for $x$ to obtain a set of $n+1$ independent linear equations for the $n+1$ variables $b_k$. Such a system has only a single solution, and it is obvious that $b_k=a_k$ is a solution.

*If you consider them as functions acting on some other field, then the claim may be false. For example in $\mathbb Z/2\mathbb Z$, $x=x^2=x^3=x^4=\ldots$, therefore as functions in that field, all you can say is that $a_0 = b_0$ (obtained by inserting $0$) and $\sum_{k=0}^n a_k = \sum_{k=0}^n b_k$ (obtained by inserting $1$).

*If you consider $x$ not as variable taking arbitrary values, but as a single specific value, then the claim is definitely false. Rather, you've got just a single equation that the coefficients have to fulfill.
