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