Divide using synthetic division $$ ( 4x^{4}-12x^{3}-8x^{2}+27x-15 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&4&-12&-8&27&-15\\& & 12& 0& -24& \color{black}{9} \\ \hline &\color{blue}{4}&\color{blue}{0}&\color{blue}{-8}&\color{blue}{3}&\color{orangered}{-6} \end{array} $$The solution is:
$$ \frac{ 4x^{4}-12x^{3}-8x^{2}+27x-15 }{ x-3 } = \color{blue}{4x^{3}-8x+3} \color{red}{~-~} \frac{ \color{red}{ 6 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&-12&-8&27&-15\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 4 }&-12&-8&27&-15\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&-12&-8&27&-15\\& & \color{blue}{12} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 12 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}3&4&\color{orangered}{ -12 }&-8&27&-15\\& & \color{orangered}{12} & & & \\ \hline &4&\color{orangered}{0}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&-12&-8&27&-15\\& & 12& \color{blue}{0} & & \\ \hline &4&\color{blue}{0}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 0 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}3&4&-12&\color{orangered}{ -8 }&27&-15\\& & 12& \color{orangered}{0} & & \\ \hline &4&0&\color{orangered}{-8}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&-12&-8&27&-15\\& & 12& 0& \color{blue}{-24} & \\ \hline &4&0&\color{blue}{-8}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 27 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}3&4&-12&-8&\color{orangered}{ 27 }&-15\\& & 12& 0& \color{orangered}{-24} & \\ \hline &4&0&-8&\color{orangered}{3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&-12&-8&27&-15\\& & 12& 0& -24& \color{blue}{9} \\ \hline &4&0&-8&\color{blue}{3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 9 } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}3&4&-12&-8&27&\color{orangered}{ -15 }\\& & 12& 0& -24& \color{orangered}{9} \\ \hline &\color{blue}{4}&\color{blue}{0}&\color{blue}{-8}&\color{blue}{3}&\color{orangered}{-6} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}-8x+3 } $ with a remainder of $ \color{red}{ -6 } $.