Divide using synthetic division $$ ( -x^{4}+7x^{3}+2x-12 ) \div ( x-7 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}7&-1&7&0&2&-12\\& & -7& 0& 0& \color{black}{14} \\ \hline &\color{blue}{-1}&\color{blue}{0}&\color{blue}{0}&\color{blue}{2}&\color{orangered}{2} \end{array} $$The solution is:
$$ \frac{ -x^{4}+7x^{3}+2x-12 }{ x-7 } = \color{blue}{-x^{3}+2} ~+~ \frac{ \color{red}{ 2 } }{ x-7 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -7 = 0 $ ( $ x = \color{blue}{ 7 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&-1&7&0&2&-12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}7&\color{orangered}{ -1 }&7&0&2&-12\\& & & & & \\ \hline &\color{orangered}{-1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -7 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&-1&7&0&2&-12\\& & \color{blue}{-7} & & & \\ \hline &\color{blue}{-1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -7 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}7&-1&\color{orangered}{ 7 }&0&2&-12\\& & \color{orangered}{-7} & & & \\ \hline &-1&\color{orangered}{0}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&-1&7&0&2&-12\\& & -7& \color{blue}{0} & & \\ \hline &-1&\color{blue}{0}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}7&-1&7&\color{orangered}{ 0 }&2&-12\\& & -7& \color{orangered}{0} & & \\ \hline &-1&0&\color{orangered}{0}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&-1&7&0&2&-12\\& & -7& 0& \color{blue}{0} & \\ \hline &-1&0&\color{blue}{0}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 0 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}7&-1&7&0&\color{orangered}{ 2 }&-12\\& & -7& 0& \color{orangered}{0} & \\ \hline &-1&0&0&\color{orangered}{2}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 2 } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&-1&7&0&2&-12\\& & -7& 0& 0& \color{blue}{14} \\ \hline &-1&0&0&\color{blue}{2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 14 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}7&-1&7&0&2&\color{orangered}{ -12 }\\& & -7& 0& 0& \color{orangered}{14} \\ \hline &\color{blue}{-1}&\color{blue}{0}&\color{blue}{0}&\color{blue}{2}&\color{orangered}{2} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -x^{3}+2 } $ with a remainder of $ \color{red}{ 2 } $.