Expanding the Polynomial Using Taylor Series? Expand the polynomial
$$f(x) = x^3-2x^2-3x+5$$
In power of $(x-2)$

This might be a simple question but all we have to do to solve this question is to expand using the Taylor series with $x =2 $ , as opposed to the Maclaurin at $x=0$.
$f(x) = f(a) + \frac{f'(a)(x-a)^1}{1!}+\frac{f''(a)(x-a)^2}{2!}+ \cdot \cdot \cdot$
$f(2)= -1+1(x-2)+\frac{8(x-2)^2}{2!}+\frac{6(x-2)^3}{3!}$
$f(2) = -1 +(x-2) +4(x-2)^2+(x-2)^3$
So in the end this would be the solution correct?

Just like for the change of basis for numbers, you can use  Horner's algorithm:
divide  the polynomial by $x-2$, then  divide again the quotient by $x-2$, and so on until you obtain a constant quotient. The coefficients of the successive powers of $x-2$ are the successive remainders (including the final quotient).
Illustration: 
\begin{array}{rrrrr}
& 1 &-2 & -3 & 5 \\
+&\downarrow&2 & 0 & -6 \\
\hline
\times2 &1 & 0 &-3 & \color{red}{-1} \\
+&\downarrow & 2&4\\
\times 2&1&2&\color{red}{1}\\ \hline
+&\downarrow & 2\\
\times 2&\color{red}1 & \color{red}4
\end{array}
So the expansion of the polynomial in powers of $x-2$ is
$$f(x)=-1+(x-2)+4(x-2)^2+(x-2)^3.$$
