Divide using synthetic division $$ ( x^{4}+10x^{3}-11x^{2}+34 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&1&10&-11&0&34\\& & -4& -24& 140& \color{black}{-560} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{-35}&\color{blue}{140}&\color{orangered}{-526} \end{array} $$The solution is:
$$ \frac{ x^{4}+10x^{3}-11x^{2}+34 }{ x+4 } = \color{blue}{x^{3}+6x^{2}-35x+140} \color{red}{~-~} \frac{ \color{red}{ 526 } }{ x+4 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&10&-11&0&34\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 1 }&10&-11&0&34\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&10&-11&0&34\\& & \color{blue}{-4} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}-4&1&\color{orangered}{ 10 }&-11&0&34\\& & \color{orangered}{-4} & & & \\ \hline &1&\color{orangered}{6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 6 } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&10&-11&0&34\\& & -4& \color{blue}{-24} & & \\ \hline &1&\color{blue}{6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -35 } $
$$ \begin{array}{c|rrrrr}-4&1&10&\color{orangered}{ -11 }&0&34\\& & -4& \color{orangered}{-24} & & \\ \hline &1&6&\color{orangered}{-35}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -35 \right) } = \color{blue}{ 140 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&10&-11&0&34\\& & -4& -24& \color{blue}{140} & \\ \hline &1&6&\color{blue}{-35}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 140 } = \color{orangered}{ 140 } $
$$ \begin{array}{c|rrrrr}-4&1&10&-11&\color{orangered}{ 0 }&34\\& & -4& -24& \color{orangered}{140} & \\ \hline &1&6&-35&\color{orangered}{140}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 140 } = \color{blue}{ -560 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&10&-11&0&34\\& & -4& -24& 140& \color{blue}{-560} \\ \hline &1&6&-35&\color{blue}{140}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 34 } + \color{orangered}{ \left( -560 \right) } = \color{orangered}{ -526 } $
$$ \begin{array}{c|rrrrr}-4&1&10&-11&0&\color{orangered}{ 34 }\\& & -4& -24& 140& \color{orangered}{-560} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{-35}&\color{blue}{140}&\color{orangered}{-526} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+6x^{2}-35x+140 } $ with a remainder of $ \color{red}{ -526 } $.