Divide using synthetic division $$ ( x^{3}-17x+32 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&1&0&-17&32\\& & -5& 25& \color{black}{-40} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{8}&\color{orangered}{-8} \end{array} $$The solution is:
$$ \frac{ x^{3}-17x+32 }{ x+5 } = \color{blue}{x^{2}-5x+8} \color{red}{~-~} \frac{ \color{red}{ 8 } }{ x+5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 5 = 0 $ ( $ x = \color{blue}{ -5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&0&-17&32\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 1 }&0&-17&32\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 1 } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&0&-17&32\\& & \color{blue}{-5} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}-5&1&\color{orangered}{ 0 }&-17&32\\& & \color{orangered}{-5} & & \\ \hline &1&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 25 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&0&-17&32\\& & -5& \color{blue}{25} & \\ \hline &1&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ 25 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}-5&1&0&\color{orangered}{ -17 }&32\\& & -5& \color{orangered}{25} & \\ \hline &1&-5&\color{orangered}{8}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 8 } = \color{blue}{ -40 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&0&-17&32\\& & -5& 25& \color{blue}{-40} \\ \hline &1&-5&\color{blue}{8}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 32 } + \color{orangered}{ \left( -40 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}-5&1&0&-17&\color{orangered}{ 32 }\\& & -5& 25& \color{orangered}{-40} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{8}&\color{orangered}{-8} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-5x+8 } $ with a remainder of $ \color{red}{ -8 } $.