Divide using synthetic division $$ ( 12x^{4}-20x^{3}-45x^{2}+80x-12 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&12&-20&-45&80&-12\\& & 36& 48& 9& \color{black}{267} \\ \hline &\color{blue}{12}&\color{blue}{16}&\color{blue}{3}&\color{blue}{89}&\color{orangered}{255} \end{array} $$The solution is:
$$ \frac{ 12x^{4}-20x^{3}-45x^{2}+80x-12 }{ x-3 } = \color{blue}{12x^{3}+16x^{2}+3x+89} ~+~ \frac{ \color{red}{ 255 } }{ x-3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&12&-20&-45&80&-12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 12 }&-20&-45&80&-12\\& & & & & \\ \hline &\color{orangered}{12}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 12 } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&12&-20&-45&80&-12\\& & \color{blue}{36} & & & \\ \hline &\color{blue}{12}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 36 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrrr}3&12&\color{orangered}{ -20 }&-45&80&-12\\& & \color{orangered}{36} & & & \\ \hline &12&\color{orangered}{16}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 16 } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&12&-20&-45&80&-12\\& & 36& \color{blue}{48} & & \\ \hline &12&\color{blue}{16}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -45 } + \color{orangered}{ 48 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}3&12&-20&\color{orangered}{ -45 }&80&-12\\& & 36& \color{orangered}{48} & & \\ \hline &12&16&\color{orangered}{3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&12&-20&-45&80&-12\\& & 36& 48& \color{blue}{9} & \\ \hline &12&16&\color{blue}{3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 80 } + \color{orangered}{ 9 } = \color{orangered}{ 89 } $
$$ \begin{array}{c|rrrrr}3&12&-20&-45&\color{orangered}{ 80 }&-12\\& & 36& 48& \color{orangered}{9} & \\ \hline &12&16&3&\color{orangered}{89}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 89 } = \color{blue}{ 267 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&12&-20&-45&80&-12\\& & 36& 48& 9& \color{blue}{267} \\ \hline &12&16&3&\color{blue}{89}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 267 } = \color{orangered}{ 255 } $
$$ \begin{array}{c|rrrrr}3&12&-20&-45&80&\color{orangered}{ -12 }\\& & 36& 48& 9& \color{orangered}{267} \\ \hline &\color{blue}{12}&\color{blue}{16}&\color{blue}{3}&\color{blue}{89}&\color{orangered}{255} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 12x^{3}+16x^{2}+3x+89 } $ with a remainder of $ \color{red}{ 255 } $.