Divide using synthetic division $$ ( 6x^{4}-10x-3 ) \div ( 2x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&6&0&0&-10&-3\\& & 12& 24& 48& \color{black}{76} \\ \hline &\color{blue}{6}&\color{blue}{12}&\color{blue}{24}&\color{blue}{38}&\color{orangered}{73} \end{array} $$The solution is:
$$ \frac{ 6x^{4}-10x-3 }{ x-2 } = \color{blue}{6x^{3}+12x^{2}+24x+38} ~+~ \frac{ \color{red}{ 73 } }{ x-2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&6&0&0&-10&-3\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 6 }&0&0&-10&-3\\& & & & & \\ \hline &\color{orangered}{6}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 6 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&6&0&0&-10&-3\\& & \color{blue}{12} & & & \\ \hline &\color{blue}{6}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 12 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}2&6&\color{orangered}{ 0 }&0&-10&-3\\& & \color{orangered}{12} & & & \\ \hline &6&\color{orangered}{12}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 12 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&6&0&0&-10&-3\\& & 12& \color{blue}{24} & & \\ \hline &6&\color{blue}{12}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 24 } = \color{orangered}{ 24 } $
$$ \begin{array}{c|rrrrr}2&6&0&\color{orangered}{ 0 }&-10&-3\\& & 12& \color{orangered}{24} & & \\ \hline &6&12&\color{orangered}{24}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 24 } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&6&0&0&-10&-3\\& & 12& 24& \color{blue}{48} & \\ \hline &6&12&\color{blue}{24}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 48 } = \color{orangered}{ 38 } $
$$ \begin{array}{c|rrrrr}2&6&0&0&\color{orangered}{ -10 }&-3\\& & 12& 24& \color{orangered}{48} & \\ \hline &6&12&24&\color{orangered}{38}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 38 } = \color{blue}{ 76 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&6&0&0&-10&-3\\& & 12& 24& 48& \color{blue}{76} \\ \hline &6&12&24&\color{blue}{38}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 76 } = \color{orangered}{ 73 } $
$$ \begin{array}{c|rrrrr}2&6&0&0&-10&\color{orangered}{ -3 }\\& & 12& 24& 48& \color{orangered}{76} \\ \hline &\color{blue}{6}&\color{blue}{12}&\color{blue}{24}&\color{blue}{38}&\color{orangered}{73} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{3}+12x^{2}+24x+38 } $ with a remainder of $ \color{red}{ 73 } $.