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