Divide using synthetic division $$ ( x^{3}+2x^{2}+x-5 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&1&2&1&-5\\& & 4& 24& \color{black}{100} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{25}&\color{orangered}{95} \end{array} $$The solution is:
$$ \frac{ x^{3}+2x^{2}+x-5 }{ x-4 } = \color{blue}{x^{2}+6x+25} ~+~ \frac{ \color{red}{ 95 } }{ x-4 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&2&1&-5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 1 }&2&1&-5\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&2&1&-5\\& & \color{blue}{4} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 4 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}4&1&\color{orangered}{ 2 }&1&-5\\& & \color{orangered}{4} & & \\ \hline &1&\color{orangered}{6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 6 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&2&1&-5\\& & 4& \color{blue}{24} & \\ \hline &1&\color{blue}{6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 24 } = \color{orangered}{ 25 } $
$$ \begin{array}{c|rrrr}4&1&2&\color{orangered}{ 1 }&-5\\& & 4& \color{orangered}{24} & \\ \hline &1&6&\color{orangered}{25}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 25 } = \color{blue}{ 100 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&2&1&-5\\& & 4& 24& \color{blue}{100} \\ \hline &1&6&\color{blue}{25}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 100 } = \color{orangered}{ 95 } $
$$ \begin{array}{c|rrrr}4&1&2&1&\color{orangered}{ -5 }\\& & 4& 24& \color{orangered}{100} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{25}&\color{orangered}{95} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+6x+25 } $ with a remainder of $ \color{red}{ 95 } $.