Question

Fin

Answer 1

Synthetic division for arbitrary \(k\) yields the following table.\[\small\begin{array}{c|cccccc}
k&7&12&6&-5&16\\
&&7k&7k^2+12k&7k^3+12k^2+6k&7k^4+12k^3+6k^2-6k\\
\hline
&7&7k+12&7k^2+12k+6&7k^3+12k^2+6k-5&\color{red}{7k^4+12k^3+6k^2-5k+16}
\end{array}\]The remainder is the red term. What's the value when \(k=3\)?

Answer 2

@Visarika23

Answer 3

just substitute k in x i.e. find f(k)
whatever u will get will be the remainder