Divide using synthetic division $$ ( 3x^{4}-16x^{3}-7x^{2}+64x-20 ) \div ( x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-1&3&-16&-7&64&-20\\& & -3& 19& -12& \color{black}{-52} \\ \hline &\color{blue}{3}&\color{blue}{-19}&\color{blue}{12}&\color{blue}{52}&\color{orangered}{-72} \end{array} $$The solution is:
$$ \frac{ 3x^{4}-16x^{3}-7x^{2}+64x-20 }{ x+1 } = \color{blue}{3x^{3}-19x^{2}+12x+52} \color{red}{~-~} \frac{ \color{red}{ 72 } }{ x+1 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&3&-16&-7&64&-20\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-1&\color{orangered}{ 3 }&-16&-7&64&-20\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 3 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&3&-16&-7&64&-20\\& & \color{blue}{-3} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -19 } $
$$ \begin{array}{c|rrrrr}-1&3&\color{orangered}{ -16 }&-7&64&-20\\& & \color{orangered}{-3} & & & \\ \hline &3&\color{orangered}{-19}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -19 \right) } = \color{blue}{ 19 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&3&-16&-7&64&-20\\& & -3& \color{blue}{19} & & \\ \hline &3&\color{blue}{-19}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 19 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}-1&3&-16&\color{orangered}{ -7 }&64&-20\\& & -3& \color{orangered}{19} & & \\ \hline &3&-19&\color{orangered}{12}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 12 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&3&-16&-7&64&-20\\& & -3& 19& \color{blue}{-12} & \\ \hline &3&-19&\color{blue}{12}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 64 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 52 } $
$$ \begin{array}{c|rrrrr}-1&3&-16&-7&\color{orangered}{ 64 }&-20\\& & -3& 19& \color{orangered}{-12} & \\ \hline &3&-19&12&\color{orangered}{52}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 52 } = \color{blue}{ -52 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&3&-16&-7&64&-20\\& & -3& 19& -12& \color{blue}{-52} \\ \hline &3&-19&12&\color{blue}{52}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ \left( -52 \right) } = \color{orangered}{ -72 } $
$$ \begin{array}{c|rrrrr}-1&3&-16&-7&64&\color{orangered}{ -20 }\\& & -3& 19& -12& \color{orangered}{-52} \\ \hline &\color{blue}{3}&\color{blue}{-19}&\color{blue}{12}&\color{blue}{52}&\color{orangered}{-72} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}-19x^{2}+12x+52 } $ with a remainder of $ \color{red}{ -72 } $.