Divide using synthetic division $$ ( 2x^{3}-2x^{2}-x+3 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&2&-2&-1&3\\& & 8& 24& \color{black}{92} \\ \hline &\color{blue}{2}&\color{blue}{6}&\color{blue}{23}&\color{orangered}{95} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-2x^{2}-x+3 }{ x-4 } = \color{blue}{2x^{2}+6x+23} ~+~ \frac{ \color{red}{ 95 } }{ x-4 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{4}&2&-2&-1&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 2 }&-2&-1&3\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 2 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&2&-2&-1&3\\& & \color{blue}{8} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 8 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}4&2&\color{orangered}{ -2 }&-1&3\\& & \color{orangered}{8} & & \\ \hline &2&\color{orangered}{6}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&2&-2&-1&3\\& & 8& \color{blue}{24} & \\ \hline &2&\color{blue}{6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 24 } = \color{orangered}{ 23 } $
$$ \begin{array}{c|rrrr}4&2&-2&\color{orangered}{ -1 }&3\\& & 8& \color{orangered}{24} & \\ \hline &2&6&\color{orangered}{23}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 23 } = \color{blue}{ 92 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&2&-2&-1&3\\& & 8& 24& \color{blue}{92} \\ \hline &2&6&\color{blue}{23}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 92 } = \color{orangered}{ 95 } $
$$ \begin{array}{c|rrrr}4&2&-2&-1&\color{orangered}{ 3 }\\& & 8& 24& \color{orangered}{92} \\ \hline &\color{blue}{2}&\color{blue}{6}&\color{blue}{23}&\color{orangered}{95} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}+6x+23 } $ with a remainder of $ \color{red}{ 95 } $.