Divide using synthetic division $$ ( 5x^{3}+2x-7 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&5&0&2&-7\\& & -25& 125& \color{black}{-635} \\ \hline &\color{blue}{5}&\color{blue}{-25}&\color{blue}{127}&\color{orangered}{-642} \end{array} $$The solution is:
$$ \frac{ 5x^{3}+2x-7 }{ x+5 } = \color{blue}{5x^{2}-25x+127} \color{red}{~-~} \frac{ \color{red}{ 642 } }{ x+5 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&5&0&2&-7\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 5 }&0&2&-7\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 5 } = \color{blue}{ -25 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&5&0&2&-7\\& & \color{blue}{-25} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -25 \right) } = \color{orangered}{ -25 } $
$$ \begin{array}{c|rrrr}-5&5&\color{orangered}{ 0 }&2&-7\\& & \color{orangered}{-25} & & \\ \hline &5&\color{orangered}{-25}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -25 \right) } = \color{blue}{ 125 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&5&0&2&-7\\& & -25& \color{blue}{125} & \\ \hline &5&\color{blue}{-25}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 125 } = \color{orangered}{ 127 } $
$$ \begin{array}{c|rrrr}-5&5&0&\color{orangered}{ 2 }&-7\\& & -25& \color{orangered}{125} & \\ \hline &5&-25&\color{orangered}{127}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 127 } = \color{blue}{ -635 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&5&0&2&-7\\& & -25& 125& \color{blue}{-635} \\ \hline &5&-25&\color{blue}{127}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ \left( -635 \right) } = \color{orangered}{ -642 } $
$$ \begin{array}{c|rrrr}-5&5&0&2&\color{orangered}{ -7 }\\& & -25& 125& \color{orangered}{-635} \\ \hline &\color{blue}{5}&\color{blue}{-25}&\color{blue}{127}&\color{orangered}{-642} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{2}-25x+127 } $ with a remainder of $ \color{red}{ -642 } $.