Divide using synthetic division $$ ( x^{4}-4x^{3}+13x^{2}+28x-140 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&1&-4&13&28&-140\\& & -4& 32& -180& \color{black}{608} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{45}&\color{blue}{-152}&\color{orangered}{468} \end{array} $$The solution is:
$$ \frac{ x^{4}-4x^{3}+13x^{2}+28x-140 }{ x+4 } = \color{blue}{x^{3}-8x^{2}+45x-152} ~+~ \frac{ \color{red}{ 468 } }{ x+4 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&-4&13&28&-140\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 1 }&-4&13&28&-140\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&-4&13&28&-140\\& & \color{blue}{-4} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}-4&1&\color{orangered}{ -4 }&13&28&-140\\& & \color{orangered}{-4} & & & \\ \hline &1&\color{orangered}{-8}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&-4&13&28&-140\\& & -4& \color{blue}{32} & & \\ \hline &1&\color{blue}{-8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ 32 } = \color{orangered}{ 45 } $
$$ \begin{array}{c|rrrrr}-4&1&-4&\color{orangered}{ 13 }&28&-140\\& & -4& \color{orangered}{32} & & \\ \hline &1&-8&\color{orangered}{45}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 45 } = \color{blue}{ -180 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&-4&13&28&-140\\& & -4& 32& \color{blue}{-180} & \\ \hline &1&-8&\color{blue}{45}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 28 } + \color{orangered}{ \left( -180 \right) } = \color{orangered}{ -152 } $
$$ \begin{array}{c|rrrrr}-4&1&-4&13&\color{orangered}{ 28 }&-140\\& & -4& 32& \color{orangered}{-180} & \\ \hline &1&-8&45&\color{orangered}{-152}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -152 \right) } = \color{blue}{ 608 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&-4&13&28&-140\\& & -4& 32& -180& \color{blue}{608} \\ \hline &1&-8&45&\color{blue}{-152}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -140 } + \color{orangered}{ 608 } = \color{orangered}{ 468 } $
$$ \begin{array}{c|rrrrr}-4&1&-4&13&28&\color{orangered}{ -140 }\\& & -4& 32& -180& \color{orangered}{608} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{45}&\color{blue}{-152}&\color{orangered}{468} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-8x^{2}+45x-152 } $ with a remainder of $ \color{red}{ 468 } $.