Divide using synthetic division $$ ( 7x^{3}-1 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&7&0&0&-1\\& & -14& 28& \color{black}{-56} \\ \hline &\color{blue}{7}&\color{blue}{-14}&\color{blue}{28}&\color{orangered}{-57} \end{array} $$The solution is:
$$ \frac{ 7x^{3}-1 }{ x+2 } = \color{blue}{7x^{2}-14x+28} \color{red}{~-~} \frac{ \color{red}{ 57 } }{ x+2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&7&0&0&-1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 7 }&0&0&-1\\& & & & \\ \hline &\color{orangered}{7}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 7 } = \color{blue}{ -14 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&7&0&0&-1\\& & \color{blue}{-14} & & \\ \hline &\color{blue}{7}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -14 \right) } = \color{orangered}{ -14 } $
$$ \begin{array}{c|rrrr}-2&7&\color{orangered}{ 0 }&0&-1\\& & \color{orangered}{-14} & & \\ \hline &7&\color{orangered}{-14}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -14 \right) } = \color{blue}{ 28 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&7&0&0&-1\\& & -14& \color{blue}{28} & \\ \hline &7&\color{blue}{-14}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 28 } = \color{orangered}{ 28 } $
$$ \begin{array}{c|rrrr}-2&7&0&\color{orangered}{ 0 }&-1\\& & -14& \color{orangered}{28} & \\ \hline &7&-14&\color{orangered}{28}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 28 } = \color{blue}{ -56 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&7&0&0&-1\\& & -14& 28& \color{blue}{-56} \\ \hline &7&-14&\color{blue}{28}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -56 \right) } = \color{orangered}{ -57 } $
$$ \begin{array}{c|rrrr}-2&7&0&0&\color{orangered}{ -1 }\\& & -14& 28& \color{orangered}{-56} \\ \hline &\color{blue}{7}&\color{blue}{-14}&\color{blue}{28}&\color{orangered}{-57} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{2}-14x+28 } $ with a remainder of $ \color{red}{ -57 } $.