Writing polynomial using powers of (x-a) I have a question related to writing a polynomial using powers of binomial of form $(x-a).$ 
I found an example: polynomial $P(x) = x^4 + 2x^3-3x^2-4x+1$ can be written as 
$ (x+1)^4-2(x+1)^3-3(x+1)^2+4(x+1)+1$ using powers of $(x+1)$ and Horner's Method. How do we obtain this representation of polynomial? How is Horner's Method used for that?
 A: As you know you may get the result by the Taylor polynomial of $$ P(x) = x^4 + 2x^3-3x^2-4x+1$$ about $x=-1$
If you insist on the Horner's  Method then it is be successive division by $x+1$
That is $$P(x) = P(-1) + P'(-1)(x+1) + P''(-1)(x+1)^2/2 +... $$
As you notice, the remainder in dividing your polynomial by $(x+1)$ is the constant which is $1$
A: You want an expression $a_4(x+1)^4+a_3(x+1)^3+a_2(x+1)^2+a_1(x+1)+a_0$. Repeatedly apply the method for $x=-1$. The last number in each row will give you the next coefficient starting from the lowest power.
\begin{array}{|c|c|c|c|c|c|}
\hline
& 1 & 2 & -3 & -4 & 1 \\ \hline
-1 & 1& 1& -4 & 0 & 1 \rightarrow a_0\\ \hline
-1 & 1 & 0& -4 & 4 \rightarrow  a_1\\ \hline
-1 & 1 & -1& -3 \rightarrow a_2\\ \hline
\end{array}
and so on.
A: This is similar to changing base of a number system.
So you can use long division.
$$P(x) = x^4 + 2x^3-3x^2-4x+1$$
To express it as powers of $(x+1)$, divide $P(x)$ by the new base $(x+1)$
$$\begin{align}
x^4 + 2x^3-3x^2-4x+1 &= (x+1)(x^3+x^2-4x) + \color{blue}{1} \\
x^3+x^2-4x &= (x+1)(x^2-4)+ \color{blue}{4}\\  
x^2-4 &= (x+1)(x-1) \color{blue}{-3}\\  
x-1 &= (x+1)(1)  \color{blue}{- 2}\\  
1 &= (x+1)0+  \color{blue}{1}\\
\end{align}$$ 
$$P(x) = \color{blue}{1}(x+1)^4 \color{blue}{-2}(x+1)^3\color{blue}{-3}(x+1)^2+\color{blue}{4}(x+1)+\color{blue}{1}$$ 
