Teorema Sisa

\(f(x)\) dibagi \((x + 1)\) memiliki sisa \(-12\) maka \(f(-1) = -12\)

\begin{equation*}
\begin{split}
f(x) & = ax^3 - 5x^2 + bx - 1 \\\\
f(-1) & = a(-1)^3 -5(-1)^2 + b(-1) - 1 \\\\
-12 & = -a -5 -b -1 \\\\
a + b & = 6 \quad {\color {red} \dotso \: (1)}
\end{split}
\end{equation*}

\(f(x)\) dibagi \((x - 2)\) memiliki sisa \(3\) maka \(f(2) = 3\)

\begin{equation*}
\begin{split}
f(x) & = ax^3 - 5x^2 + bx - 1 \\\\
f(2) & = a(2)^3 -5(2)^2 + b(2) - 1 \\\\
3 & = 8a -20 +2b -1 \\\\
8a + 2b & = 24 \\\\
4a + b & = 12 \quad {\color {red} \dotso \: (2)}
\end{split}
\end{equation*}

Eliminasi persamaan (1) dan (2)

\begin{equation*}
\begin{split}
4a + b & = 12 \\\\
a + b & = 6 \quad(-) \\
\hline\\
3a & = 6 \\\\
a & = 2
\end{split}
\end{equation*}

Substitusi a = 2 ke persamaan (1)

\begin{equation*}
\begin{split}
a + b & = 6 \\\\
2 + b & = 6 \\\\
b & = 4
\end{split}
\end{equation*}