Divide using synthetic division $$ ( x^{4}+3x^{3}-14x^{2}+15 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&1&3&-14&0&15\\& & 3& 18& 12& \color{black}{36} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{4}&\color{blue}{12}&\color{orangered}{51} \end{array} $$The solution is:
$$ \frac{ x^{4}+3x^{3}-14x^{2}+15 }{ x-3 } = \color{blue}{x^{3}+6x^{2}+4x+12} ~+~ \frac{ \color{red}{ 51 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&3&-14&0&15\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 1 }&3&-14&0&15\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&3&-14&0&15\\& & \color{blue}{3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 3 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}3&1&\color{orangered}{ 3 }&-14&0&15\\& & \color{orangered}{3} & & & \\ \hline &1&\color{orangered}{6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 6 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&3&-14&0&15\\& & 3& \color{blue}{18} & & \\ \hline &1&\color{blue}{6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 18 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}3&1&3&\color{orangered}{ -14 }&0&15\\& & 3& \color{orangered}{18} & & \\ \hline &1&6&\color{orangered}{4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&3&-14&0&15\\& & 3& 18& \color{blue}{12} & \\ \hline &1&6&\color{blue}{4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 12 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}3&1&3&-14&\color{orangered}{ 0 }&15\\& & 3& 18& \color{orangered}{12} & \\ \hline &1&6&4&\color{orangered}{12}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 12 } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&3&-14&0&15\\& & 3& 18& 12& \color{blue}{36} \\ \hline &1&6&4&\color{blue}{12}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ 36 } = \color{orangered}{ 51 } $
$$ \begin{array}{c|rrrrr}3&1&3&-14&0&\color{orangered}{ 15 }\\& & 3& 18& 12& \color{orangered}{36} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{4}&\color{blue}{12}&\color{orangered}{51} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+6x^{2}+4x+12 } $ with a remainder of $ \color{red}{ 51 } $.