Divide using synthetic division $$ ( 6x^{4}+19x^{3}-4x+15 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&6&19&0&-4&15\\& & -18& -3& 9& \color{black}{-15} \\ \hline &\color{blue}{6}&\color{blue}{1}&\color{blue}{-3}&\color{blue}{5}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 6x^{4}+19x^{3}-4x+15 }{ x+3 } = \color{blue}{6x^{3}+x^{2}-3x+5} $$Explanation
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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&6&19&0&-4&15\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 6 }&19&0&-4&15\\& & & & & \\ \hline &\color{orangered}{6}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 6 } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&6&19&0&-4&15\\& & \color{blue}{-18} & & & \\ \hline &\color{blue}{6}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}-3&6&\color{orangered}{ 19 }&0&-4&15\\& & \color{orangered}{-18} & & & \\ \hline &6&\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&6&19&0&-4&15\\& & -18& \color{blue}{-3} & & \\ \hline &6&\color{blue}{1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-3&6&19&\color{orangered}{ 0 }&-4&15\\& & -18& \color{orangered}{-3} & & \\ \hline &6&1&\color{orangered}{-3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&6&19&0&-4&15\\& & -18& -3& \color{blue}{9} & \\ \hline &6&1&\color{blue}{-3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 9 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}-3&6&19&0&\color{orangered}{ -4 }&15\\& & -18& -3& \color{orangered}{9} & \\ \hline &6&1&-3&\color{orangered}{5}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 5 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&6&19&0&-4&15\\& & -18& -3& 9& \color{blue}{-15} \\ \hline &6&1&-3&\color{blue}{5}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-3&6&19&0&-4&\color{orangered}{ 15 }\\& & -18& -3& 9& \color{orangered}{-15} \\ \hline &\color{blue}{6}&\color{blue}{1}&\color{blue}{-3}&\color{blue}{5}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{3}+x^{2}-3x+5 } $ with a remainder of $ \color{red}{ 0 } $.