Divide using synthetic division $$ ( x^{3}+11x^{2}+20x-32 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&1&11&20&-32\\& & 4& 60& \color{black}{320} \\ \hline &\color{blue}{1}&\color{blue}{15}&\color{blue}{80}&\color{orangered}{288} \end{array} $$The solution is:
$$ \frac{ x^{3}+11x^{2}+20x-32 }{ x-4 } = \color{blue}{x^{2}+15x+80} ~+~ \frac{ \color{red}{ 288 } }{ x-4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&11&20&-32\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 1 }&11&20&-32\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 1 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&11&20&-32\\& & \color{blue}{4} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ 4 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrr}4&1&\color{orangered}{ 11 }&20&-32\\& & \color{orangered}{4} & & \\ \hline &1&\color{orangered}{15}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 15 } = \color{blue}{ 60 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&11&20&-32\\& & 4& \color{blue}{60} & \\ \hline &1&\color{blue}{15}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ 60 } = \color{orangered}{ 80 } $
$$ \begin{array}{c|rrrr}4&1&11&\color{orangered}{ 20 }&-32\\& & 4& \color{orangered}{60} & \\ \hline &1&15&\color{orangered}{80}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 80 } = \color{blue}{ 320 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&11&20&-32\\& & 4& 60& \color{blue}{320} \\ \hline &1&15&\color{blue}{80}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -32 } + \color{orangered}{ 320 } = \color{orangered}{ 288 } $
$$ \begin{array}{c|rrrr}4&1&11&20&\color{orangered}{ -32 }\\& & 4& 60& \color{orangered}{320} \\ \hline &\color{blue}{1}&\color{blue}{15}&\color{blue}{80}&\color{orangered}{288} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+15x+80 } $ with a remainder of $ \color{red}{ 288 } $.