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