Divide using synthetic division $$ ( 3x^{4}-7x^{3}-14x^{2}+15x+27 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&3&-7&-14&15&27\\& & 9& 6& -24& \color{black}{-27} \\ \hline &\color{blue}{3}&\color{blue}{2}&\color{blue}{-8}&\color{blue}{-9}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 3x^{4}-7x^{3}-14x^{2}+15x+27 }{ x-3 } = \color{blue}{3x^{3}+2x^{2}-8x-9} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&3&-7&-14&15&27\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 3 }&-7&-14&15&27\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&3&-7&-14&15&27\\& & \color{blue}{9} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 9 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}3&3&\color{orangered}{ -7 }&-14&15&27\\& & \color{orangered}{9} & & & \\ \hline &3&\color{orangered}{2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&3&-7&-14&15&27\\& & 9& \color{blue}{6} & & \\ \hline &3&\color{blue}{2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 6 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}3&3&-7&\color{orangered}{ -14 }&15&27\\& & 9& \color{orangered}{6} & & \\ \hline &3&2&\color{orangered}{-8}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&3&-7&-14&15&27\\& & 9& 6& \color{blue}{-24} & \\ \hline &3&2&\color{blue}{-8}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrrr}3&3&-7&-14&\color{orangered}{ 15 }&27\\& & 9& 6& \color{orangered}{-24} & \\ \hline &3&2&-8&\color{orangered}{-9}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ -27 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&3&-7&-14&15&27\\& & 9& 6& -24& \color{blue}{-27} \\ \hline &3&2&-8&\color{blue}{-9}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 27 } + \color{orangered}{ \left( -27 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}3&3&-7&-14&15&\color{orangered}{ 27 }\\& & 9& 6& -24& \color{orangered}{-27} \\ \hline &\color{blue}{3}&\color{blue}{2}&\color{blue}{-8}&\color{blue}{-9}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}+2x^{2}-8x-9 } $ with a remainder of $ \color{red}{ 0 } $.