Divide using synthetic division $$ ( 3x^{4}-16x^{3}-7x^{2}+64x-20 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&3&-16&-7&64&-20\\& & 3& -13& -20& \color{black}{44} \\ \hline &\color{blue}{3}&\color{blue}{-13}&\color{blue}{-20}&\color{blue}{44}&\color{orangered}{24} \end{array} $$The solution is:
$$ \frac{ 3x^{4}-16x^{3}-7x^{2}+64x-20 }{ x-1 } = \color{blue}{3x^{3}-13x^{2}-20x+44} ~+~ \frac{ \color{red}{ 24 } }{ x-1 } $$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}{ 3 } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrrr}1&3&\color{orangered}{ -16 }&-7&64&-20\\& & \color{orangered}{3} & & & \\ \hline &3&\color{orangered}{-13}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -13 \right) } = \color{blue}{ -13 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&3&-16&-7&64&-20\\& & 3& \color{blue}{-13} & & \\ \hline &3&\color{blue}{-13}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ \left( -13 \right) } = \color{orangered}{ -20 } $
$$ \begin{array}{c|rrrrr}1&3&-16&\color{orangered}{ -7 }&64&-20\\& & 3& \color{orangered}{-13} & & \\ \hline &3&-13&\color{orangered}{-20}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -20 \right) } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&3&-16&-7&64&-20\\& & 3& -13& \color{blue}{-20} & \\ \hline &3&-13&\color{blue}{-20}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 64 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ 44 } $
$$ \begin{array}{c|rrrrr}1&3&-16&-7&\color{orangered}{ 64 }&-20\\& & 3& -13& \color{orangered}{-20} & \\ \hline &3&-13&-20&\color{orangered}{44}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 44 } = \color{blue}{ 44 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&3&-16&-7&64&-20\\& & 3& -13& -20& \color{blue}{44} \\ \hline &3&-13&-20&\color{blue}{44}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 44 } = \color{orangered}{ 24 } $
$$ \begin{array}{c|rrrrr}1&3&-16&-7&64&\color{orangered}{ -20 }\\& & 3& -13& -20& \color{orangered}{44} \\ \hline &\color{blue}{3}&\color{blue}{-13}&\color{blue}{-20}&\color{blue}{44}&\color{orangered}{24} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}-13x^{2}-20x+44 } $ with a remainder of $ \color{red}{ 24 } $.