Divide using synthetic division $$ ( 5x^{4}+12x^{3}-21x^{2}-40x-12 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&5&12&-21&-40&-12\\& & 15& 81& 180& \color{black}{420} \\ \hline &\color{blue}{5}&\color{blue}{27}&\color{blue}{60}&\color{blue}{140}&\color{orangered}{408} \end{array} $$The solution is:
$$ \frac{ 5x^{4}+12x^{3}-21x^{2}-40x-12 }{ x-3 } = \color{blue}{5x^{3}+27x^{2}+60x+140} ~+~ \frac{ \color{red}{ 408 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&5&12&-21&-40&-12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 5 }&12&-21&-40&-12\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 5 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&5&12&-21&-40&-12\\& & \color{blue}{15} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 15 } = \color{orangered}{ 27 } $
$$ \begin{array}{c|rrrrr}3&5&\color{orangered}{ 12 }&-21&-40&-12\\& & \color{orangered}{15} & & & \\ \hline &5&\color{orangered}{27}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 27 } = \color{blue}{ 81 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&5&12&-21&-40&-12\\& & 15& \color{blue}{81} & & \\ \hline &5&\color{blue}{27}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -21 } + \color{orangered}{ 81 } = \color{orangered}{ 60 } $
$$ \begin{array}{c|rrrrr}3&5&12&\color{orangered}{ -21 }&-40&-12\\& & 15& \color{orangered}{81} & & \\ \hline &5&27&\color{orangered}{60}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 60 } = \color{blue}{ 180 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&5&12&-21&-40&-12\\& & 15& 81& \color{blue}{180} & \\ \hline &5&27&\color{blue}{60}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -40 } + \color{orangered}{ 180 } = \color{orangered}{ 140 } $
$$ \begin{array}{c|rrrrr}3&5&12&-21&\color{orangered}{ -40 }&-12\\& & 15& 81& \color{orangered}{180} & \\ \hline &5&27&60&\color{orangered}{140}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 140 } = \color{blue}{ 420 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&5&12&-21&-40&-12\\& & 15& 81& 180& \color{blue}{420} \\ \hline &5&27&60&\color{blue}{140}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 420 } = \color{orangered}{ 408 } $
$$ \begin{array}{c|rrrrr}3&5&12&-21&-40&\color{orangered}{ -12 }\\& & 15& 81& 180& \color{orangered}{420} \\ \hline &\color{blue}{5}&\color{blue}{27}&\color{blue}{60}&\color{blue}{140}&\color{orangered}{408} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{3}+27x^{2}+60x+140 } $ with a remainder of $ \color{red}{ 408 } $.