Divide using synthetic division $$ ( x^{4}-7x^{3}+14x^{2}-38x-60 ) \div ( 3x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-1&1&-7&14&-38&-60\\& & -1& 8& -22& \color{black}{60} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{22}&\color{blue}{-60}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{4}-7x^{3}+14x^{2}-38x-60 }{ x+1 } = \color{blue}{x^{3}-8x^{2}+22x-60} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&-7&14&-38&-60\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-1&\color{orangered}{ 1 }&-7&14&-38&-60\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 1 } = \color{blue}{ -1 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&-7&14&-38&-60\\& & \color{blue}{-1} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}-1&1&\color{orangered}{ -7 }&14&-38&-60\\& & \color{orangered}{-1} & & & \\ \hline &1&\color{orangered}{-8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&-7&14&-38&-60\\& & -1& \color{blue}{8} & & \\ \hline &1&\color{blue}{-8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ 8 } = \color{orangered}{ 22 } $
$$ \begin{array}{c|rrrrr}-1&1&-7&\color{orangered}{ 14 }&-38&-60\\& & -1& \color{orangered}{8} & & \\ \hline &1&-8&\color{orangered}{22}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 22 } = \color{blue}{ -22 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&-7&14&-38&-60\\& & -1& 8& \color{blue}{-22} & \\ \hline &1&-8&\color{blue}{22}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -38 } + \color{orangered}{ \left( -22 \right) } = \color{orangered}{ -60 } $
$$ \begin{array}{c|rrrrr}-1&1&-7&14&\color{orangered}{ -38 }&-60\\& & -1& 8& \color{orangered}{-22} & \\ \hline &1&-8&22&\color{orangered}{-60}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -60 \right) } = \color{blue}{ 60 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&-7&14&-38&-60\\& & -1& 8& -22& \color{blue}{60} \\ \hline &1&-8&22&\color{blue}{-60}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -60 } + \color{orangered}{ 60 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-1&1&-7&14&-38&\color{orangered}{ -60 }\\& & -1& 8& -22& \color{orangered}{60} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{22}&\color{blue}{-60}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-8x^{2}+22x-60 } $ with a remainder of $ \color{red}{ 0 } $.