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