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