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Note. Please provide only a hint along with some explanation, but not the answer. I want to struggle with this problem. This is not homework.

Show that for any number $c$, a polynomial $ P(x) = b_0 + b_1 x + b_2 x^2 + \cdots b_n x^n$ can also be written as $P(x) = a_0 + a_1 (x - c) + a_2 ( x- c)^2 + a_3(x - c)^3\cdots a_n(x - c)^n$ where $a_0 = P(c)$.

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Have you tried induction? – user38268 Dec 26 '12 at 12:45
@BenjaLim: How will I do a proof by induction here? – Parth Kohli Dec 26 '12 at 12:46
Perhaps consider $P(x+c)$? – Daniel Littlewood Dec 26 '12 at 12:49
$= b_0 + b_1(x + c) + b_2(x + c)^2 +\cdots b_n(x + c)^n$ – Parth Kohli Dec 26 '12 at 13:19
up vote 3 down vote accepted

$$x^r=(x-c+c)^r$$ $$=\sum_{0\le t\le r}\binom rt (x-c)^{r-t}c^t=(x-c)^r+\binom r1 (x-c)^{r-1}c+ \binom r2 (x-c)^{r-2}c^2+\cdots+\binom r{r-2} (x-c)^{2}c^{r-2}+\binom r{r-1} (x-c)^1c^{r-1}+c^r$$

Now put $r=0,1,2,3,\cdots,n-1,n$

Use the fact $\binom nr=0$ if $r>n$ or $r<0$

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Ah, took me a while to understand this one. Thanks @lab :) – Parth Kohli Dec 26 '12 at 13:28
@DumbCow, I did not mention the Binomial Theorem by name,, as well-known.Anyway welcome. – lab bhattacharjee Dec 26 '12 at 18:00

Let $Q(x) = P(x + c)$. So $Q(x - c) = P(x)$. What is $Q(0)$? Can you show that $Q(x)$ is a polynomial?

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How will that help? I mean, it is not really proven that both are equivalent... – Parth Kohli Dec 26 '12 at 13:14

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