Divide using synthetic division $$ ( -x^{4}+4x^{3}+23x+10 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&-1&4&0&23&10\\& & -5& -5& -25& \color{black}{-10} \\ \hline &\color{blue}{-1}&\color{blue}{-1}&\color{blue}{-5}&\color{blue}{-2}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ -x^{4}+4x^{3}+23x+10 }{ x-5 } = \color{blue}{-x^{3}-x^{2}-5x-2} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&-1&4&0&23&10\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ -1 }&4&0&23&10\\& & & & & \\ \hline &\color{orangered}{-1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&-1&4&0&23&10\\& & \color{blue}{-5} & & & \\ \hline &\color{blue}{-1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}5&-1&\color{orangered}{ 4 }&0&23&10\\& & \color{orangered}{-5} & & & \\ \hline &-1&\color{orangered}{-1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&-1&4&0&23&10\\& & -5& \color{blue}{-5} & & \\ \hline &-1&\color{blue}{-1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}5&-1&4&\color{orangered}{ 0 }&23&10\\& & -5& \color{orangered}{-5} & & \\ \hline &-1&-1&\color{orangered}{-5}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -25 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&-1&4&0&23&10\\& & -5& -5& \color{blue}{-25} & \\ \hline &-1&-1&\color{blue}{-5}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 23 } + \color{orangered}{ \left( -25 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}5&-1&4&0&\color{orangered}{ 23 }&10\\& & -5& -5& \color{orangered}{-25} & \\ \hline &-1&-1&-5&\color{orangered}{-2}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&-1&4&0&23&10\\& & -5& -5& -25& \color{blue}{-10} \\ \hline &-1&-1&-5&\color{blue}{-2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}5&-1&4&0&23&\color{orangered}{ 10 }\\& & -5& -5& -25& \color{orangered}{-10} \\ \hline &\color{blue}{-1}&\color{blue}{-1}&\color{blue}{-5}&\color{blue}{-2}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -x^{3}-x^{2}-5x-2 } $ with a remainder of $ \color{red}{ 0 } $.