synthetic division question. Synthetic division is possible when the Divisior is in the form of $x+a$ or $x-a$. but what if the divisor is in the form of $x^2+a$, $x^2-a$, $x^3-a$,... and higher powers. how can we perform synthetic division in such cases. 
Thanks 
 A: For the particular case of divisors in the form $x^n-a$ it is possible to replace the long division with $n$ synthetic divisions.
For a polynomial $P(x)$ being divided by $x^3-a$ for example, group the powers of $x$ in $P$ according to the remainder $\bmod 3$ and write it as:
$$
P(x) = P_0(x^3) + x P_1(x^3) + x^2P_2(x^3) 
$$
Use synthetic division to calculate the quotients and remainders of the following:
$$
P_k(x) = (x-a)Q_k(x) + r_k \quad\quad \text{for} \;\; k=0,1,2
$$
Then:
$$
P(x)=(x^3-a)Q(x) + R(x)
$$
where $Q(x) = Q_0(x^3) + xQ_1(x^3)+x^2Q_2(x^3)$ and $R(x)=r_0+r_1x+r_2x^2\,$.

[ EDIT ] Following is a fully worked out example for $P(X)=x^4-6x^3+16x^2-25x+10$ (the polynomial was borrowed from another, unrelated question) being divided by $x^3-2$.


*

*Group the powers:


$$P(X)=x^4-6x^3+16x^2-25x+10= (-6x^3+10) + x\cdot (x^3-25) + x^2 \cdot 16$$
$$
\iff
\begin{cases}
\begin{align}
 P_0(x) & = -6x+10 \\
 P_1(x) & = x - 25 \\
 P_2(x) &= 16
\end{align}
\end{cases}
$$


*

*Divide $P_k$ by $x-2$ and determine $Q_k,r_k$ by synthetic division:


$$
\begin{cases}
\begin{alignat}{3}
 P_0(x) & = -6x+10 && = -6(x-2) - 2\\
 P_1(x) & = x - 25 && = (x-2) - 23\\
 P_2(x) & = 16 && = 16
\end{alignat}
\end{cases}
$$
$$
\iff
\begin{cases}
\begin{align}
 Q_0(x) & = -6 \,,\;\; r_0 = -2\\
 Q_1(x) & = 1  \,,\;\; r_1 = - 23\\
 Q_2(x) & = 0  \,,\;\; r_2 = 16
\end{align}
\end{cases}
$$


*

*Calculate $Q,R$:


$$
Q(x) = Q_0(x^3) + xQ_1(x^3)+x^2Q_2(x^3) = -6 +x+ x^2 \cdot 0 = x-6\\
R(x)=r_0+r_1x+r_2x^2=16x^2-23 x-2\,$$


*

*Verify that indeed:


$$x^4-6x^3+16x^2-25x+10=(x^3-2)(x-6)+ 16x^2-23 x -2$$
