Divide using synthetic division $$ ( 3x^{3}+4x^{2}-15x+30 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-4&3&4&-15&30\\& & -12& 32& \color{black}{-68} \\ \hline &\color{blue}{3}&\color{blue}{-8}&\color{blue}{17}&\color{orangered}{-38} \end{array} $$The solution is:
$$ \frac{ 3x^{3}+4x^{2}-15x+30 }{ x+4 } = \color{blue}{3x^{2}-8x+17} \color{red}{~-~} \frac{ \color{red}{ 38 } }{ x+4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 4 = 0 $ ( $ x = \color{blue}{ -4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&3&4&-15&30\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-4&\color{orangered}{ 3 }&4&-15&30\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 3 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&3&4&-15&30\\& & \color{blue}{-12} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}-4&3&\color{orangered}{ 4 }&-15&30\\& & \color{orangered}{-12} & & \\ \hline &3&\color{orangered}{-8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&3&4&-15&30\\& & -12& \color{blue}{32} & \\ \hline &3&\color{blue}{-8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 32 } = \color{orangered}{ 17 } $
$$ \begin{array}{c|rrrr}-4&3&4&\color{orangered}{ -15 }&30\\& & -12& \color{orangered}{32} & \\ \hline &3&-8&\color{orangered}{17}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 17 } = \color{blue}{ -68 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&3&4&-15&30\\& & -12& 32& \color{blue}{-68} \\ \hline &3&-8&\color{blue}{17}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 30 } + \color{orangered}{ \left( -68 \right) } = \color{orangered}{ -38 } $
$$ \begin{array}{c|rrrr}-4&3&4&-15&\color{orangered}{ 30 }\\& & -12& 32& \color{orangered}{-68} \\ \hline &\color{blue}{3}&\color{blue}{-8}&\color{blue}{17}&\color{orangered}{-38} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}-8x+17 } $ with a remainder of $ \color{red}{ -38 } $.