Divide using synthetic division $$ ( x^{4}+2x^{3}-5x^{2}+8x+12 ) \div ( x-1 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&1&2&-5&8&12\\& & 1& 3& -2& \color{black}{6} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{-2}&\color{blue}{6}&\color{orangered}{18} \end{array} $$The solution is:
$$ \dfrac{ x^{4}+2x^{3}-5x^{2}+8x+12 }{ x-1 } = \color{blue}{x^{3}+3x^{2}-2x+6} ~+~ \dfrac{ \color{red}{ 18 } }{ x-1 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&2&-5&8&12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 1 }&2&-5&8&12\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 1 } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&2&-5&8&12\\& & \color{blue}{1} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 1 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}1&1&\color{orangered}{ 2 }&-5&8&12\\& & \color{orangered}{1} & & & \\ \hline &1&\color{orangered}{3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 3 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&2&-5&8&12\\& & 1& \color{blue}{3} & & \\ \hline &1&\color{blue}{3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 3 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}1&1&2&\color{orangered}{ -5 }&8&12\\& & 1& \color{orangered}{3} & & \\ \hline &1&3&\color{orangered}{-2}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&2&-5&8&12\\& & 1& 3& \color{blue}{-2} & \\ \hline &1&3&\color{blue}{-2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}1&1&2&-5&\color{orangered}{ 8 }&12\\& & 1& 3& \color{orangered}{-2} & \\ \hline &1&3&-2&\color{orangered}{6}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 6 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&2&-5&8&12\\& & 1& 3& -2& \color{blue}{6} \\ \hline &1&3&-2&\color{blue}{6}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 6 } = \color{orangered}{ 18 } $
$$ \begin{array}{c|rrrrr}1&1&2&-5&8&\color{orangered}{ 12 }\\& & 1& 3& -2& \color{orangered}{6} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{-2}&\color{blue}{6}&\color{orangered}{18} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+3x^{2}-2x+6 } $ with a remainder of $ \color{red}{ 18 } $.