Divide using synthetic division $$ ( x^{4}-4x^{3}+x-10 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&1&-4&0&1&-10\\& & -4& 32& -128& \color{black}{508} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{32}&\color{blue}{-127}&\color{orangered}{498} \end{array} $$The solution is:
$$ \frac{ x^{4}-4x^{3}+x-10 }{ x+4 } = \color{blue}{x^{3}-8x^{2}+32x-127} ~+~ \frac{ \color{red}{ 498 } }{ x+4 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&-4&0&1&-10\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 1 }&-4&0&1&-10\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&-4&0&1&-10\\& & \color{blue}{-4} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}-4&1&\color{orangered}{ -4 }&0&1&-10\\& & \color{orangered}{-4} & & & \\ \hline &1&\color{orangered}{-8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&-4&0&1&-10\\& & -4& \color{blue}{32} & & \\ \hline &1&\color{blue}{-8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 32 } = \color{orangered}{ 32 } $
$$ \begin{array}{c|rrrrr}-4&1&-4&\color{orangered}{ 0 }&1&-10\\& & -4& \color{orangered}{32} & & \\ \hline &1&-8&\color{orangered}{32}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 32 } = \color{blue}{ -128 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&-4&0&1&-10\\& & -4& 32& \color{blue}{-128} & \\ \hline &1&-8&\color{blue}{32}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -128 \right) } = \color{orangered}{ -127 } $
$$ \begin{array}{c|rrrrr}-4&1&-4&0&\color{orangered}{ 1 }&-10\\& & -4& 32& \color{orangered}{-128} & \\ \hline &1&-8&32&\color{orangered}{-127}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -127 \right) } = \color{blue}{ 508 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&-4&0&1&-10\\& & -4& 32& -128& \color{blue}{508} \\ \hline &1&-8&32&\color{blue}{-127}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 508 } = \color{orangered}{ 498 } $
$$ \begin{array}{c|rrrrr}-4&1&-4&0&1&\color{orangered}{ -10 }\\& & -4& 32& -128& \color{orangered}{508} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{32}&\color{blue}{-127}&\color{orangered}{498} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-8x^{2}+32x-127 } $ with a remainder of $ \color{red}{ 498 } $.