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?)


  • $\begingroup$ The coefficients don’t really disappear, they just get renamed. $\endgroup$ – ViHdzP Dec 9 '19 at 8:55

The $a_i$'s are what you get after the expansion. If, for instance, $P(x)=b_0+b_1x+b_2x^2$, then\begin{align}P(y)&=b_0+b_1(x+c)+b_2(x+c)^2\\&=b_0+b_1x-b_1c+b_2x^2-2b_2cx+b_2c^2\\&=\overbrace{b_0+b_1c+b_2c^2}^{\phantom{a_0}=a_0}+\overbrace{(b_1+2b_2c)}^{\phantom{a_1}=a_1}x+\overbrace{b_2}^{\phantom{a_2}=a_2}x^2.\end{align}And, as in this case, you always have $a_n=b_n$

  • $\begingroup$ Thanks for you answer! The set of $y=x+c$ Shouldn't it have been $y=x-c$? $\endgroup$ – epcode Dec 9 '19 at 9:09
  • $\begingroup$ The sutitution $y=x+c$ Shouldn't it have been $y=x-c$? I do not understand the change of signs from P(y) to P(x) $\endgroup$ – epcode Dec 9 '19 at 9:22
  • $\begingroup$ It was a typo. I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Dec 9 '19 at 14:32

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