Using division, express the polynomial $z^3+2$ in the form: $$a_0+a_1(z-1)+a_2(z-1)^2+a_3(z-1)^3.$$
Typically, when I need to divide polynomials, I look for a common factor, but I don't see any in this polynomial. My strategy would be to treat this as a linear algebra problem and expand each quantity to solve for each $a_n$ term. Would that be necessary in this case or is there a simpler way to go about this?
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For $f(z)=z^{3}+2$, then $f(z)=f(1)+f'(1)(z-1)+\dfrac{f''(1)}{2!}(z-1)^{2}+\dfrac{f^{(3)}(1)}{3!}(z-1)^{3}$.
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$$z^3+2=a_0+a_1(z-1)+a_2(z-1)^2+a_3(z-1)^3.$$
$$ z=1 \implies a_0 =3$$ Differentiate, you get $$3z^2 =a_1+2a_2 (z-1)+ 3a_3(z-1)^2$$
$$ z=1 \implies a_1=3$$
Differentiate, you get
$$ 6z=2a_2+6a_3(z-1)$$
$$ z=1 \implies a_2=3$$
Differentiate, you get
$$ a_3 =1$$
Thus $$z^3+2=3+3(z-1)+3(z-1)^2+(z-1)^3.$$
As suggested in comments, other methods are probably easier, but you certainly can do this by division if you wish. We want $$z^3+2=a_0+a_1(z-1)+a_2(z-1)^2+a_3(z-1)^3\ .\tag{$*$}$$ Dividing the LHS by $z-1$ gives $$z^3+2=(z-1)(z^2+z+1)+3\ .$$ Looking at $(*)$, when we divide the RHS by $z-1$ the remainder is obviously $a_0$; for the LHS it is $3$ as we have just shown; so $a_0=3$. Now in $(*)$, subtract $3$ from both sides and divide by $z-1$; we have already done the working so we just write down $$z^2+z+1=a_1+a_2(z-1)+a_3(z-1)^2\ .$$ Now follow the same procedure to find $a_1$, and so on.