Divide using synthetic division $$ ( 6x^{3}-15x^{2}-19x+37 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&6&-15&-19&37\\& & 18& 9& \color{black}{-30} \\ \hline &\color{blue}{6}&\color{blue}{3}&\color{blue}{-10}&\color{orangered}{7} \end{array} $$The solution is:
$$ \frac{ 6x^{3}-15x^{2}-19x+37 }{ x-3 } = \color{blue}{6x^{2}+3x-10} ~+~ \frac{ \color{red}{ 7 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&6&-15&-19&37\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 6 }&-15&-19&37\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&6&-15&-19&37\\& & \color{blue}{18} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 18 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}3&6&\color{orangered}{ -15 }&-19&37\\& & \color{orangered}{18} & & \\ \hline &6&\color{orangered}{3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&6&-15&-19&37\\& & 18& \color{blue}{9} & \\ \hline &6&\color{blue}{3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -19 } + \color{orangered}{ 9 } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrr}3&6&-15&\color{orangered}{ -19 }&37\\& & 18& \color{orangered}{9} & \\ \hline &6&3&\color{orangered}{-10}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ -30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&6&-15&-19&37\\& & 18& 9& \color{blue}{-30} \\ \hline &6&3&\color{blue}{-10}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 37 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}3&6&-15&-19&\color{orangered}{ 37 }\\& & 18& 9& \color{orangered}{-30} \\ \hline &\color{blue}{6}&\color{blue}{3}&\color{blue}{-10}&\color{orangered}{7} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}+3x-10 } $ with a remainder of $ \color{red}{ 7 } $.