Divide using synthetic division $$ ( 4x^{4}+21x^{3}-26x^{2}+28x-10 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-5&4&21&-26&28&-10\\& & -20& -5& 155& \color{black}{-915} \\ \hline &\color{blue}{4}&\color{blue}{1}&\color{blue}{-31}&\color{blue}{183}&\color{orangered}{-925} \end{array} $$The solution is:
$$ \frac{ 4x^{4}+21x^{3}-26x^{2}+28x-10 }{ x+5 } = \color{blue}{4x^{3}+x^{2}-31x+183} \color{red}{~-~} \frac{ \color{red}{ 925 } }{ x+5 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&4&21&-26&28&-10\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-5&\color{orangered}{ 4 }&21&-26&28&-10\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 4 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&4&21&-26&28&-10\\& & \color{blue}{-20} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}-5&4&\color{orangered}{ 21 }&-26&28&-10\\& & \color{orangered}{-20} & & & \\ \hline &4&\color{orangered}{1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 1 } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&4&21&-26&28&-10\\& & -20& \color{blue}{-5} & & \\ \hline &4&\color{blue}{1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -26 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -31 } $
$$ \begin{array}{c|rrrrr}-5&4&21&\color{orangered}{ -26 }&28&-10\\& & -20& \color{orangered}{-5} & & \\ \hline &4&1&\color{orangered}{-31}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -31 \right) } = \color{blue}{ 155 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&4&21&-26&28&-10\\& & -20& -5& \color{blue}{155} & \\ \hline &4&1&\color{blue}{-31}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 28 } + \color{orangered}{ 155 } = \color{orangered}{ 183 } $
$$ \begin{array}{c|rrrrr}-5&4&21&-26&\color{orangered}{ 28 }&-10\\& & -20& -5& \color{orangered}{155} & \\ \hline &4&1&-31&\color{orangered}{183}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 183 } = \color{blue}{ -915 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&4&21&-26&28&-10\\& & -20& -5& 155& \color{blue}{-915} \\ \hline &4&1&-31&\color{blue}{183}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ \left( -915 \right) } = \color{orangered}{ -925 } $
$$ \begin{array}{c|rrrrr}-5&4&21&-26&28&\color{orangered}{ -10 }\\& & -20& -5& 155& \color{orangered}{-915} \\ \hline &\color{blue}{4}&\color{blue}{1}&\color{blue}{-31}&\color{blue}{183}&\color{orangered}{-925} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}+x^{2}-31x+183 } $ with a remainder of $ \color{red}{ -925 } $.