Divide using synthetic division $$ ( 7x^{3}+14x-7 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-4&7&0&14&-7\\& & -28& 112& \color{black}{-504} \\ \hline &\color{blue}{7}&\color{blue}{-28}&\color{blue}{126}&\color{orangered}{-511} \end{array} $$The solution is:
$$ \frac{ 7x^{3}+14x-7 }{ x+4 } = \color{blue}{7x^{2}-28x+126} \color{red}{~-~} \frac{ \color{red}{ 511 } }{ x+4 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&7&0&14&-7\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-4&\color{orangered}{ 7 }&0&14&-7\\& & & & \\ \hline &\color{orangered}{7}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 7 } = \color{blue}{ -28 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&7&0&14&-7\\& & \color{blue}{-28} & & \\ \hline &\color{blue}{7}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -28 \right) } = \color{orangered}{ -28 } $
$$ \begin{array}{c|rrrr}-4&7&\color{orangered}{ 0 }&14&-7\\& & \color{orangered}{-28} & & \\ \hline &7&\color{orangered}{-28}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -28 \right) } = \color{blue}{ 112 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&7&0&14&-7\\& & -28& \color{blue}{112} & \\ \hline &7&\color{blue}{-28}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ 112 } = \color{orangered}{ 126 } $
$$ \begin{array}{c|rrrr}-4&7&0&\color{orangered}{ 14 }&-7\\& & -28& \color{orangered}{112} & \\ \hline &7&-28&\color{orangered}{126}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 126 } = \color{blue}{ -504 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&7&0&14&-7\\& & -28& 112& \color{blue}{-504} \\ \hline &7&-28&\color{blue}{126}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ \left( -504 \right) } = \color{orangered}{ -511 } $
$$ \begin{array}{c|rrrr}-4&7&0&14&\color{orangered}{ -7 }\\& & -28& 112& \color{orangered}{-504} \\ \hline &\color{blue}{7}&\color{blue}{-28}&\color{blue}{126}&\color{orangered}{-511} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{2}-28x+126 } $ with a remainder of $ \color{red}{ -511 } $.