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