Divide using synthetic division $$ ( -5x^{4}-5x^{3}+x^{2}+2 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-5&-5&-5&1&0&2\\& & 25& -100& 495& \color{black}{-2475} \\ \hline &\color{blue}{-5}&\color{blue}{20}&\color{blue}{-99}&\color{blue}{495}&\color{orangered}{-2473} \end{array} $$The solution is:
$$ \frac{ -5x^{4}-5x^{3}+x^{2}+2 }{ x+5 } = \color{blue}{-5x^{3}+20x^{2}-99x+495} \color{red}{~-~} \frac{ \color{red}{ 2473 } }{ x+5 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&-5&-5&1&0&2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-5&\color{orangered}{ -5 }&-5&1&0&2\\& & & & & \\ \hline &\color{orangered}{-5}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 25 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&-5&-5&1&0&2\\& & \color{blue}{25} & & & \\ \hline &\color{blue}{-5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 25 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrrr}-5&-5&\color{orangered}{ -5 }&1&0&2\\& & \color{orangered}{25} & & & \\ \hline &-5&\color{orangered}{20}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 20 } = \color{blue}{ -100 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&-5&-5&1&0&2\\& & 25& \color{blue}{-100} & & \\ \hline &-5&\color{blue}{20}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -100 \right) } = \color{orangered}{ -99 } $
$$ \begin{array}{c|rrrrr}-5&-5&-5&\color{orangered}{ 1 }&0&2\\& & 25& \color{orangered}{-100} & & \\ \hline &-5&20&\color{orangered}{-99}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -99 \right) } = \color{blue}{ 495 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&-5&-5&1&0&2\\& & 25& -100& \color{blue}{495} & \\ \hline &-5&20&\color{blue}{-99}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 495 } = \color{orangered}{ 495 } $
$$ \begin{array}{c|rrrrr}-5&-5&-5&1&\color{orangered}{ 0 }&2\\& & 25& -100& \color{orangered}{495} & \\ \hline &-5&20&-99&\color{orangered}{495}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 495 } = \color{blue}{ -2475 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&-5&-5&1&0&2\\& & 25& -100& 495& \color{blue}{-2475} \\ \hline &-5&20&-99&\color{blue}{495}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -2475 \right) } = \color{orangered}{ -2473 } $
$$ \begin{array}{c|rrrrr}-5&-5&-5&1&0&\color{orangered}{ 2 }\\& & 25& -100& 495& \color{orangered}{-2475} \\ \hline &\color{blue}{-5}&\color{blue}{20}&\color{blue}{-99}&\color{blue}{495}&\color{orangered}{-2473} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -5x^{3}+20x^{2}-99x+495 } $ with a remainder of $ \color{red}{ -2473 } $.