Divide using synthetic division $$ ( x^{3}-11x^{2}+36x-48 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&1&-11&36&-48\\& & 5& -30& \color{black}{30} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{6}&\color{orangered}{-18} \end{array} $$The solution is:
$$ \frac{ x^{3}-11x^{2}+36x-48 }{ x-5 } = \color{blue}{x^{2}-6x+6} \color{red}{~-~} \frac{ \color{red}{ 18 } }{ x-5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-11&36&-48\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 1 }&-11&36&-48\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-11&36&-48\\& & \color{blue}{5} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 5 } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}5&1&\color{orangered}{ -11 }&36&-48\\& & \color{orangered}{5} & & \\ \hline &1&\color{orangered}{-6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-11&36&-48\\& & 5& \color{blue}{-30} & \\ \hline &1&\color{blue}{-6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 36 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}5&1&-11&\color{orangered}{ 36 }&-48\\& & 5& \color{orangered}{-30} & \\ \hline &1&-6&\color{orangered}{6}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 6 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-11&36&-48\\& & 5& -30& \color{blue}{30} \\ \hline &1&-6&\color{blue}{6}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -48 } + \color{orangered}{ 30 } = \color{orangered}{ -18 } $
$$ \begin{array}{c|rrrr}5&1&-11&36&\color{orangered}{ -48 }\\& & 5& -30& \color{orangered}{30} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{6}&\color{orangered}{-18} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-6x+6 } $ with a remainder of $ \color{red}{ -18 } $.