Divide using synthetic division $$ ( 12x^{3}-x^{2}-7x+2 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&12&-1&-7&2\\& & 24& 46& \color{black}{78} \\ \hline &\color{blue}{12}&\color{blue}{23}&\color{blue}{39}&\color{orangered}{80} \end{array} $$The solution is:
$$ \frac{ 12x^{3}-x^{2}-7x+2 }{ x-2 } = \color{blue}{12x^{2}+23x+39} ~+~ \frac{ \color{red}{ 80 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&12&-1&-7&2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 12 }&-1&-7&2\\& & & & \\ \hline &\color{orangered}{12}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 12 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&12&-1&-7&2\\& & \color{blue}{24} & & \\ \hline &\color{blue}{12}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 24 } = \color{orangered}{ 23 } $
$$ \begin{array}{c|rrrr}2&12&\color{orangered}{ -1 }&-7&2\\& & \color{orangered}{24} & & \\ \hline &12&\color{orangered}{23}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 23 } = \color{blue}{ 46 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&12&-1&-7&2\\& & 24& \color{blue}{46} & \\ \hline &12&\color{blue}{23}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 46 } = \color{orangered}{ 39 } $
$$ \begin{array}{c|rrrr}2&12&-1&\color{orangered}{ -7 }&2\\& & 24& \color{orangered}{46} & \\ \hline &12&23&\color{orangered}{39}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 39 } = \color{blue}{ 78 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&12&-1&-7&2\\& & 24& 46& \color{blue}{78} \\ \hline &12&23&\color{blue}{39}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 78 } = \color{orangered}{ 80 } $
$$ \begin{array}{c|rrrr}2&12&-1&-7&\color{orangered}{ 2 }\\& & 24& 46& \color{orangered}{78} \\ \hline &\color{blue}{12}&\color{blue}{23}&\color{blue}{39}&\color{orangered}{80} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 12x^{2}+23x+39 } $ with a remainder of $ \color{red}{ 80 } $.