Expressing $x^5-2x^3+6x^2+1$ as a sum of powers of $x+2$ I found this question in one old calculus exam on my university. It's simple enough:

Express $x^5-2x^3+6x^2+1$ as a sum of powers of $x+2$

Now, this seems like a straightforward (although slightly computationally annoying) linear algebra problem of taking the expasions of $(x+2)^p, p\in [0,5]$ and sticking the coefficients in a matrix, so I suspect I am missing some cleverer solution.
Thanks!
 A: I think that an easy way is to build a Taylor expansion around $x=-2$. As a result, you should have
$$9 + 32 (2 + x) - 62 (2 + x)^2 + 38 (2 + x)^3 - 10 (2 + x)^4 + (2 + x)^5$$
A: Here's another way:
$$\begin{array}{c|c|c|c|c|c}
      1 & 0 & -2 & 6 & 0 & 1         \\ \hline
      1 & -2 & 2 & 2 & -4 & 9       \\[0.55ex]
      1 & -4 & 10 & -18 & 32\\[0.55ex]
      1 & -6 & 22 & -62 \\[0.55ex]
      1 & -8 & 38 \\[0.55ex]
      1 & -10\\[0.55ex]
      1
\end{array}$$
Now let's look how the table is obtained. The first row represents the coefficients of the polynomial. Every number in the left-most colum is the same and has the same value as the coefficient in front of the highest degree. Now because we want to represent it as sum of powers of $(x+2)$, our approximatiozation will be at $x=-2$. So every other number is obtained in such way that we multiply the the number that's left to him by $x=-2$ and add it to the number that's above it. 
So for example for $10$ in the third row we have: $(-2) \times (-4) + 2 = 10$ and so on.
Now use the values in the main diagonal and the give polynomial will be equivalent to:
$$(x+2)^5 - 10(x+2)^4 + 38(x+2)^3 - 62(x+2)^2 + 32(x+2)^1 + 9(x+2)^0$$
This method is called Horner Method/Scheme/Division and in fact represents polynomial division.
A: Hint: Set $y=x+2$, so that $x=y-2$; do your math and …
Hint 2: Taylor expansion.
A: Hint: Taylor series around $x=-2$
Your way is good too!
A: Alternative solution: You know you need $1\cdot (x + 2)^5$, since $1$ is the coefficient in front of $x^5$. Calculate $$x^5 - 2x^3 + 6x^2 + 1 - 1\cdot(x + 2)^5$$ and deduce how many $(x + 2)^4$ you need. Rinse and repeat.
