Divide using synthetic division $$ ( 7x^{3}+2x-9 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-4&7&0&2&-9\\& & -28& 112& \color{black}{-456} \\ \hline &\color{blue}{7}&\color{blue}{-28}&\color{blue}{114}&\color{orangered}{-465} \end{array} $$The solution is:
$$ \frac{ 7x^{3}+2x-9 }{ x+4 } = \color{blue}{7x^{2}-28x+114} \color{red}{~-~} \frac{ \color{red}{ 465 } }{ x+4 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&7&0&2&-9\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-4&\color{orangered}{ 7 }&0&2&-9\\& & & & \\ \hline &\color{orangered}{7}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&7&0&2&-9\\& & \color{blue}{-28} & & \\ \hline &\color{blue}{7}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -28 \right) } = \color{orangered}{ -28 } $
$$ \begin{array}{c|rrrr}-4&7&\color{orangered}{ 0 }&2&-9\\& & \color{orangered}{-28} & & \\ \hline &7&\color{orangered}{-28}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&7&0&2&-9\\& & -28& \color{blue}{112} & \\ \hline &7&\color{blue}{-28}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 112 } = \color{orangered}{ 114 } $
$$ \begin{array}{c|rrrr}-4&7&0&\color{orangered}{ 2 }&-9\\& & -28& \color{orangered}{112} & \\ \hline &7&-28&\color{orangered}{114}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 114 } = \color{blue}{ -456 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&7&0&2&-9\\& & -28& 112& \color{blue}{-456} \\ \hline &7&-28&\color{blue}{114}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ \left( -456 \right) } = \color{orangered}{ -465 } $
$$ \begin{array}{c|rrrr}-4&7&0&2&\color{orangered}{ -9 }\\& & -28& 112& \color{orangered}{-456} \\ \hline &\color{blue}{7}&\color{blue}{-28}&\color{blue}{114}&\color{orangered}{-465} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{2}-28x+114 } $ with a remainder of $ \color{red}{ -465 } $.