Divide using synthetic division $$ ( 42x^{3}-15x^{2}+15 ) \div ( 6x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&42&-15&0&15\\& & -126& 423& \color{black}{-1269} \\ \hline &\color{blue}{42}&\color{blue}{-141}&\color{blue}{423}&\color{orangered}{-1254} \end{array} $$The solution is:
$$ \frac{ 42x^{3}-15x^{2}+15 }{ x+3 } = \color{blue}{42x^{2}-141x+423} \color{red}{~-~} \frac{ \color{red}{ 1254 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&42&-15&0&15\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 42 }&-15&0&15\\& & & & \\ \hline &\color{orangered}{42}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 42 } = \color{blue}{ -126 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&42&-15&0&15\\& & \color{blue}{-126} & & \\ \hline &\color{blue}{42}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ \left( -126 \right) } = \color{orangered}{ -141 } $
$$ \begin{array}{c|rrrr}-3&42&\color{orangered}{ -15 }&0&15\\& & \color{orangered}{-126} & & \\ \hline &42&\color{orangered}{-141}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -141 \right) } = \color{blue}{ 423 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&42&-15&0&15\\& & -126& \color{blue}{423} & \\ \hline &42&\color{blue}{-141}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 423 } = \color{orangered}{ 423 } $
$$ \begin{array}{c|rrrr}-3&42&-15&\color{orangered}{ 0 }&15\\& & -126& \color{orangered}{423} & \\ \hline &42&-141&\color{orangered}{423}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 423 } = \color{blue}{ -1269 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&42&-15&0&15\\& & -126& 423& \color{blue}{-1269} \\ \hline &42&-141&\color{blue}{423}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -1269 \right) } = \color{orangered}{ -1254 } $
$$ \begin{array}{c|rrrr}-3&42&-15&0&\color{orangered}{ 15 }\\& & -126& 423& \color{orangered}{-1269} \\ \hline &\color{blue}{42}&\color{blue}{-141}&\color{blue}{423}&\color{orangered}{-1254} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 42x^{2}-141x+423 } $ with a remainder of $ \color{red}{ -1254 } $.