I need help with this proof.
Show that for any number $c$, a polynomial $P(x)=b_0+b_1x+b_2x^2+...+b_nx^n$ can also be written $P(x)=a_0+a_1(x-c)+a_2(x-c)^2+...+a_n(x-c)^n$ where $a_0=P(c)$. Show that $a_n\ne 0$ if $b_n \ne 0$.
The solution manual indicates:
Let $y=x+c$. Then
Expanding all the binomials and collecting like terms results in a polynomial
$a_0+a_1x+a_2x^2+...+a_nx^n$ (This is where I get lost. Why the coefficients $b_i$ disappear and $a_i$ appear? How to see the expansion of binomials and the grouping of similar terms in an easy way?)
This is $P(x+c)$, so
Note that $a_n=b_n$. (Why?)