Divide using synthetic division $$ ( 3x^{3}+4x-1 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-4&3&0&4&-1\\& & -12& 48& \color{black}{-208} \\ \hline &\color{blue}{3}&\color{blue}{-12}&\color{blue}{52}&\color{orangered}{-209} \end{array} $$The solution is:
$$ \frac{ 3x^{3}+4x-1 }{ x+4 } = \color{blue}{3x^{2}-12x+52} \color{red}{~-~} \frac{ \color{red}{ 209 } }{ x+4 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&3&0&4&-1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-4&\color{orangered}{ 3 }&0&4&-1\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&3&0&4&-1\\& & \color{blue}{-12} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrr}-4&3&\color{orangered}{ 0 }&4&-1\\& & \color{orangered}{-12} & & \\ \hline &3&\color{orangered}{-12}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&3&0&4&-1\\& & -12& \color{blue}{48} & \\ \hline &3&\color{blue}{-12}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 48 } = \color{orangered}{ 52 } $
$$ \begin{array}{c|rrrr}-4&3&0&\color{orangered}{ 4 }&-1\\& & -12& \color{orangered}{48} & \\ \hline &3&-12&\color{orangered}{52}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 52 } = \color{blue}{ -208 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&3&0&4&-1\\& & -12& 48& \color{blue}{-208} \\ \hline &3&-12&\color{blue}{52}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -208 \right) } = \color{orangered}{ -209 } $
$$ \begin{array}{c|rrrr}-4&3&0&4&\color{orangered}{ -1 }\\& & -12& 48& \color{orangered}{-208} \\ \hline &\color{blue}{3}&\color{blue}{-12}&\color{blue}{52}&\color{orangered}{-209} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}-12x+52 } $ with a remainder of $ \color{red}{ -209 } $.