# Show a polynomial can also be written ...

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

$$P(y)=b_0+b_1(x+c)+b_2(x+c)^2+...+b_n(x+c)^n$$

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

$$P(x)=a_0+a_1(x-c)+a_2(x-c)^2+...+a_n(x-c)^n$$.

Note that $$a_n=b_n$$. (Why?)

Thanks.

• The coefficients don’t really disappear, they just get renamed. 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 $$a_n=b_n$$
• Thanks for you answer! The set of $y=x+c$ Shouldn't it have been $y=x-c$? Dec 9 '19 at 9:09
• 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) Dec 9 '19 at 9:22