Remainder Theorem when Divisor not linear $x-a$ Upon receiving a question, it seemed to have needed me to use the remainder theorem however for a divisor that was not linear. Now while I could long divide it (or synthetically), I was wondering how $x-a$ would apply to something like $x^2-a$ or in general $x^k-a$. This would be a great help, thank you :)

\begin{array}{c|c}
  \hline
  \text{Divisor} & \text{Remainder} \\ \hline
  x-a & f(a) \\[5pt]
  (x-a)(x-b) &
  \frac{x-b}{a-b}f(a)+\frac{x-a}{b-a}f(b) \\[5pt]
  x^2-a^2 &
  \frac{f(a)-f(-a)}{2a}x+\frac{f(a)+f(-a)}{2} \\[5pt]
  (x-a)^{2} & (x-a)f'(a)+f(a) \\[5pt]
  (x-a)(x-b)(x-c) &
  \frac{(x-b)(x-c)}{(a-b)(a-c)} f(a)+
  \frac{(x-c)(x-a)}{(b-c)(b-a)} f(b)+
  \frac{(x-a)(x-b)}{(c-a)(c-a)} f(c) \\[5pt]
  (x-a)^{2} (x-b) &  
  (x-a)^{2} \frac{f(b)-f(a)}{(b-a)^{2}}+
  \frac{(x-a)(x-b)}{a-b} f'(a)+f(a) \\[5pt]
  (x-a)^{3} & \frac{(x-a)^{2}}{2}f''(a)+(x-a)f'(a)+f(a) \\ \hline
\end{array}
