Divide using synthetic division $$ ( x^{4}+5x^{3}-15x^{2}-12x-60 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&1&5&-15&-12&-60\\& & 3& 24& 27& \color{black}{45} \\ \hline &\color{blue}{1}&\color{blue}{8}&\color{blue}{9}&\color{blue}{15}&\color{orangered}{-15} \end{array} $$The solution is:
$$ \frac{ x^{4}+5x^{3}-15x^{2}-12x-60 }{ x-3 } = \color{blue}{x^{3}+8x^{2}+9x+15} \color{red}{~-~} \frac{ \color{red}{ 15 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&5&-15&-12&-60\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 1 }&5&-15&-12&-60\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&5&-15&-12&-60\\& & \color{blue}{3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 3 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}3&1&\color{orangered}{ 5 }&-15&-12&-60\\& & \color{orangered}{3} & & & \\ \hline &1&\color{orangered}{8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 8 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&5&-15&-12&-60\\& & 3& \color{blue}{24} & & \\ \hline &1&\color{blue}{8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 24 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}3&1&5&\color{orangered}{ -15 }&-12&-60\\& & 3& \color{orangered}{24} & & \\ \hline &1&8&\color{orangered}{9}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 9 } = \color{blue}{ 27 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&5&-15&-12&-60\\& & 3& 24& \color{blue}{27} & \\ \hline &1&8&\color{blue}{9}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 27 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrrr}3&1&5&-15&\color{orangered}{ -12 }&-60\\& & 3& 24& \color{orangered}{27} & \\ \hline &1&8&9&\color{orangered}{15}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 15 } = \color{blue}{ 45 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&5&-15&-12&-60\\& & 3& 24& 27& \color{blue}{45} \\ \hline &1&8&9&\color{blue}{15}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -60 } + \color{orangered}{ 45 } = \color{orangered}{ -15 } $
$$ \begin{array}{c|rrrrr}3&1&5&-15&-12&\color{orangered}{ -60 }\\& & 3& 24& 27& \color{orangered}{45} \\ \hline &\color{blue}{1}&\color{blue}{8}&\color{blue}{9}&\color{blue}{15}&\color{orangered}{-15} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+8x^{2}+9x+15 } $ with a remainder of $ \color{red}{ -15 } $.