Divide using synthetic division $$ ( x^{5}+7x^{4}+10x^{3}-15x-22 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-3&1&7&10&0&-15&-22\\& & -3& -12& 6& -18& \color{black}{99} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{-2}&\color{blue}{6}&\color{blue}{-33}&\color{orangered}{77} \end{array} $$The solution is:
$$ \frac{ x^{5}+7x^{4}+10x^{3}-15x-22 }{ x+3 } = \color{blue}{x^{4}+4x^{3}-2x^{2}+6x-33} ~+~ \frac{ \color{red}{ 77 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&7&10&0&-15&-22\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-3&\color{orangered}{ 1 }&7&10&0&-15&-22\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&7&10&0&-15&-22\\& & \color{blue}{-3} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrrr}-3&1&\color{orangered}{ 7 }&10&0&-15&-22\\& & \color{orangered}{-3} & & & & \\ \hline &1&\color{orangered}{4}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 4 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&7&10&0&-15&-22\\& & -3& \color{blue}{-12} & & & \\ \hline &1&\color{blue}{4}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}-3&1&7&\color{orangered}{ 10 }&0&-15&-22\\& & -3& \color{orangered}{-12} & & & \\ \hline &1&4&\color{orangered}{-2}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&7&10&0&-15&-22\\& & -3& -12& \color{blue}{6} & & \\ \hline &1&4&\color{blue}{-2}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 6 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrrr}-3&1&7&10&\color{orangered}{ 0 }&-15&-22\\& & -3& -12& \color{orangered}{6} & & \\ \hline &1&4&-2&\color{orangered}{6}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 6 } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&7&10&0&-15&-22\\& & -3& -12& 6& \color{blue}{-18} & \\ \hline &1&4&-2&\color{blue}{6}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -33 } $
$$ \begin{array}{c|rrrrrr}-3&1&7&10&0&\color{orangered}{ -15 }&-22\\& & -3& -12& 6& \color{orangered}{-18} & \\ \hline &1&4&-2&6&\color{orangered}{-33}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -33 \right) } = \color{blue}{ 99 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&7&10&0&-15&-22\\& & -3& -12& 6& -18& \color{blue}{99} \\ \hline &1&4&-2&6&\color{blue}{-33}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -22 } + \color{orangered}{ 99 } = \color{orangered}{ 77 } $
$$ \begin{array}{c|rrrrrr}-3&1&7&10&0&-15&\color{orangered}{ -22 }\\& & -3& -12& 6& -18& \color{orangered}{99} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{-2}&\color{blue}{6}&\color{blue}{-33}&\color{orangered}{77} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}+4x^{3}-2x^{2}+6x-33 } $ with a remainder of $ \color{red}{ 77 } $.