Divide using synthetic division $$ ( 2x^{4}-12x^{3}+17x^{2}-7x+12 ) \div ( x-4 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&2&-12&17&-7&12\\& & 8& -16& 4& \color{black}{-12} \\ \hline &\color{blue}{2}&\color{blue}{-4}&\color{blue}{1}&\color{blue}{-3}&\color{orangered}{0} \end{array} $$The solution is:
$$ \dfrac{ 2x^{4}-12x^{3}+17x^{2}-7x+12 }{ x-4 } = \color{blue}{2x^{3}-4x^{2}+x-3} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-12&17&-7&12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}4&\color{orangered}{ 2 }&-12&17&-7&12\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-12&17&-7&12\\& & \color{blue}{8} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 8 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}4&2&\color{orangered}{ -12 }&17&-7&12\\& & \color{orangered}{8} & & & \\ \hline &2&\color{orangered}{-4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-12&17&-7&12\\& & 8& \color{blue}{-16} & & \\ \hline &2&\color{blue}{-4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 17 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}4&2&-12&\color{orangered}{ 17 }&-7&12\\& & 8& \color{orangered}{-16} & & \\ \hline &2&-4&\color{orangered}{1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 1 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-12&17&-7&12\\& & 8& -16& \color{blue}{4} & \\ \hline &2&-4&\color{blue}{1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 4 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}4&2&-12&17&\color{orangered}{ -7 }&12\\& & 8& -16& \color{orangered}{4} & \\ \hline &2&-4&1&\color{orangered}{-3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-12&17&-7&12\\& & 8& -16& 4& \color{blue}{-12} \\ \hline &2&-4&1&\color{blue}{-3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}4&2&-12&17&-7&\color{orangered}{ 12 }\\& & 8& -16& 4& \color{orangered}{-12} \\ \hline &\color{blue}{2}&\color{blue}{-4}&\color{blue}{1}&\color{blue}{-3}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-4x^{2}+x-3 } $ with a remainder of $ \color{red}{ 0 } $.