Divide using synthetic division $$ ( 3x^{4}-16x^{3}-7x^{2}+64x-20 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&3&-16&-7&64&-20\\& & 15& -5& -60& \color{black}{20} \\ \hline &\color{blue}{3}&\color{blue}{-1}&\color{blue}{-12}&\color{blue}{4}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 3x^{4}-16x^{3}-7x^{2}+64x-20 }{ x-5 } = \color{blue}{3x^{3}-x^{2}-12x+4} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&-16&-7&64&-20\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 3 }&-16&-7&64&-20\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 3 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&-16&-7&64&-20\\& & \color{blue}{15} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 15 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}5&3&\color{orangered}{ -16 }&-7&64&-20\\& & \color{orangered}{15} & & & \\ \hline &3&\color{orangered}{-1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&-16&-7&64&-20\\& & 15& \color{blue}{-5} & & \\ \hline &3&\color{blue}{-1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrr}5&3&-16&\color{orangered}{ -7 }&64&-20\\& & 15& \color{orangered}{-5} & & \\ \hline &3&-1&\color{orangered}{-12}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ -60 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&-16&-7&64&-20\\& & 15& -5& \color{blue}{-60} & \\ \hline &3&-1&\color{blue}{-12}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 64 } + \color{orangered}{ \left( -60 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}5&3&-16&-7&\color{orangered}{ 64 }&-20\\& & 15& -5& \color{orangered}{-60} & \\ \hline &3&-1&-12&\color{orangered}{4}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 4 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&-16&-7&64&-20\\& & 15& -5& -60& \color{blue}{20} \\ \hline &3&-1&-12&\color{blue}{4}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 20 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}5&3&-16&-7&64&\color{orangered}{ -20 }\\& & 15& -5& -60& \color{orangered}{20} \\ \hline &\color{blue}{3}&\color{blue}{-1}&\color{blue}{-12}&\color{blue}{4}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}-x^{2}-12x+4 } $ with a remainder of $ \color{red}{ 0 } $.