Divide using synthetic division $$ ( 12x^{3}+21x^{2}-61x+20 ) \div ( 4x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&12&21&-61&20\\& & 60& 405& \color{black}{1720} \\ \hline &\color{blue}{12}&\color{blue}{81}&\color{blue}{344}&\color{orangered}{1740} \end{array} $$The solution is:
$$ \frac{ 12x^{3}+21x^{2}-61x+20 }{ x-5 } = \color{blue}{12x^{2}+81x+344} ~+~ \frac{ \color{red}{ 1740 } }{ x-5 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{5}&12&21&-61&20\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 12 }&21&-61&20\\& & & & \\ \hline &\color{orangered}{12}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 12 } = \color{blue}{ 60 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&12&21&-61&20\\& & \color{blue}{60} & & \\ \hline &\color{blue}{12}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ 60 } = \color{orangered}{ 81 } $
$$ \begin{array}{c|rrrr}5&12&\color{orangered}{ 21 }&-61&20\\& & \color{orangered}{60} & & \\ \hline &12&\color{orangered}{81}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 81 } = \color{blue}{ 405 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&12&21&-61&20\\& & 60& \color{blue}{405} & \\ \hline &12&\color{blue}{81}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -61 } + \color{orangered}{ 405 } = \color{orangered}{ 344 } $
$$ \begin{array}{c|rrrr}5&12&21&\color{orangered}{ -61 }&20\\& & 60& \color{orangered}{405} & \\ \hline &12&81&\color{orangered}{344}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 344 } = \color{blue}{ 1720 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&12&21&-61&20\\& & 60& 405& \color{blue}{1720} \\ \hline &12&81&\color{blue}{344}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ 1720 } = \color{orangered}{ 1740 } $
$$ \begin{array}{c|rrrr}5&12&21&-61&\color{orangered}{ 20 }\\& & 60& 405& \color{orangered}{1720} \\ \hline &\color{blue}{12}&\color{blue}{81}&\color{blue}{344}&\color{orangered}{1740} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 12x^{2}+81x+344 } $ with a remainder of $ \color{red}{ 1740 } $.