Divide using synthetic division $$ ( 7x^{3}+14x^{2}-26x-9 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&7&14&-26&-9\\& & -21& 21& \color{black}{15} \\ \hline &\color{blue}{7}&\color{blue}{-7}&\color{blue}{-5}&\color{orangered}{6} \end{array} $$The solution is:
$$ \frac{ 7x^{3}+14x^{2}-26x-9 }{ x+3 } = \color{blue}{7x^{2}-7x-5} ~+~ \frac{ \color{red}{ 6 } }{ x+3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&7&14&-26&-9\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 7 }&14&-26&-9\\& & & & \\ \hline &\color{orangered}{7}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 7 } = \color{blue}{ -21 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&7&14&-26&-9\\& & \color{blue}{-21} & & \\ \hline &\color{blue}{7}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ \left( -21 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}-3&7&\color{orangered}{ 14 }&-26&-9\\& & \color{orangered}{-21} & & \\ \hline &7&\color{orangered}{-7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 21 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&7&14&-26&-9\\& & -21& \color{blue}{21} & \\ \hline &7&\color{blue}{-7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -26 } + \color{orangered}{ 21 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}-3&7&14&\color{orangered}{ -26 }&-9\\& & -21& \color{orangered}{21} & \\ \hline &7&-7&\color{orangered}{-5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&7&14&-26&-9\\& & -21& 21& \color{blue}{15} \\ \hline &7&-7&\color{blue}{-5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 15 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}-3&7&14&-26&\color{orangered}{ -9 }\\& & -21& 21& \color{orangered}{15} \\ \hline &\color{blue}{7}&\color{blue}{-7}&\color{blue}{-5}&\color{orangered}{6} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{2}-7x-5 } $ with a remainder of $ \color{red}{ 6 } $.