Divide using synthetic division $$ ( x^{5}-5x^{3}+8x^{2}-14x-15 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-3&1&0&-5&8&-14&-15\\& & -3& 9& -12& 12& \color{black}{6} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{4}&\color{blue}{-4}&\color{blue}{-2}&\color{orangered}{-9} \end{array} $$The solution is:
$$ \frac{ x^{5}-5x^{3}+8x^{2}-14x-15 }{ x+3 } = \color{blue}{x^{4}-3x^{3}+4x^{2}-4x-2} \color{red}{~-~} \frac{ \color{red}{ 9 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&0&-5&8&-14&-15\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-3&\color{orangered}{ 1 }&0&-5&8&-14&-15\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&0&-5&8&-14&-15\\& & \color{blue}{-3} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrrr}-3&1&\color{orangered}{ 0 }&-5&8&-14&-15\\& & \color{orangered}{-3} & & & & \\ \hline &1&\color{orangered}{-3}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&0&-5&8&-14&-15\\& & -3& \color{blue}{9} & & & \\ \hline &1&\color{blue}{-3}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 9 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrrr}-3&1&0&\color{orangered}{ -5 }&8&-14&-15\\& & -3& \color{orangered}{9} & & & \\ \hline &1&-3&\color{orangered}{4}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 4 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&0&-5&8&-14&-15\\& & -3& 9& \color{blue}{-12} & & \\ \hline &1&-3&\color{blue}{4}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrrr}-3&1&0&-5&\color{orangered}{ 8 }&-14&-15\\& & -3& 9& \color{orangered}{-12} & & \\ \hline &1&-3&4&\color{orangered}{-4}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&0&-5&8&-14&-15\\& & -3& 9& -12& \color{blue}{12} & \\ \hline &1&-3&4&\color{blue}{-4}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 12 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}-3&1&0&-5&8&\color{orangered}{ -14 }&-15\\& & -3& 9& -12& \color{orangered}{12} & \\ \hline &1&-3&4&-4&\color{orangered}{-2}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&0&-5&8&-14&-15\\& & -3& 9& -12& 12& \color{blue}{6} \\ \hline &1&-3&4&-4&\color{blue}{-2}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 6 } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrrrr}-3&1&0&-5&8&-14&\color{orangered}{ -15 }\\& & -3& 9& -12& 12& \color{orangered}{6} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{4}&\color{blue}{-4}&\color{blue}{-2}&\color{orangered}{-9} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}-3x^{3}+4x^{2}-4x-2 } $ with a remainder of $ \color{red}{ -9 } $.