Divide using synthetic division $$ ( 8x^{4}-14x^{3}+x^{2}-12x+12 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&8&-14&1&-12&12\\& & 16& 4& 10& \color{black}{-4} \\ \hline &\color{blue}{8}&\color{blue}{2}&\color{blue}{5}&\color{blue}{-2}&\color{orangered}{8} \end{array} $$The solution is:
$$ \frac{ 8x^{4}-14x^{3}+x^{2}-12x+12 }{ x-2 } = \color{blue}{8x^{3}+2x^{2}+5x-2} ~+~ \frac{ \color{red}{ 8 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&8&-14&1&-12&12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 8 }&-14&1&-12&12\\& & & & & \\ \hline &\color{orangered}{8}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 8 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&8&-14&1&-12&12\\& & \color{blue}{16} & & & \\ \hline &\color{blue}{8}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 16 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}2&8&\color{orangered}{ -14 }&1&-12&12\\& & \color{orangered}{16} & & & \\ \hline &8&\color{orangered}{2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&8&-14&1&-12&12\\& & 16& \color{blue}{4} & & \\ \hline &8&\color{blue}{2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 4 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}2&8&-14&\color{orangered}{ 1 }&-12&12\\& & 16& \color{orangered}{4} & & \\ \hline &8&2&\color{orangered}{5}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 5 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&8&-14&1&-12&12\\& & 16& 4& \color{blue}{10} & \\ \hline &8&2&\color{blue}{5}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 10 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}2&8&-14&1&\color{orangered}{ -12 }&12\\& & 16& 4& \color{orangered}{10} & \\ \hline &8&2&5&\color{orangered}{-2}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&8&-14&1&-12&12\\& & 16& 4& 10& \color{blue}{-4} \\ \hline &8&2&5&\color{blue}{-2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}2&8&-14&1&-12&\color{orangered}{ 12 }\\& & 16& 4& 10& \color{orangered}{-4} \\ \hline &\color{blue}{8}&\color{blue}{2}&\color{blue}{5}&\color{blue}{-2}&\color{orangered}{8} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 8x^{3}+2x^{2}+5x-2 } $ with a remainder of $ \color{red}{ 8 } $.