Divide using synthetic division $$ ( x^{4}-5x^{3}+8x^{2}+15x-2 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&1&-5&8&15&-2\\& & 3& -6& 6& \color{black}{63} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{2}&\color{blue}{21}&\color{orangered}{61} \end{array} $$The solution is:
$$ \frac{ x^{4}-5x^{3}+8x^{2}+15x-2 }{ x-3 } = \color{blue}{x^{3}-2x^{2}+2x+21} ~+~ \frac{ \color{red}{ 61 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-5&8&15&-2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 1 }&-5&8&15&-2\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-5&8&15&-2\\& & \color{blue}{3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 3 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}3&1&\color{orangered}{ -5 }&8&15&-2\\& & \color{orangered}{3} & & & \\ \hline &1&\color{orangered}{-2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-5&8&15&-2\\& & 3& \color{blue}{-6} & & \\ \hline &1&\color{blue}{-2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}3&1&-5&\color{orangered}{ 8 }&15&-2\\& & 3& \color{orangered}{-6} & & \\ \hline &1&-2&\color{orangered}{2}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-5&8&15&-2\\& & 3& -6& \color{blue}{6} & \\ \hline &1&-2&\color{blue}{2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ 6 } = \color{orangered}{ 21 } $
$$ \begin{array}{c|rrrrr}3&1&-5&8&\color{orangered}{ 15 }&-2\\& & 3& -6& \color{orangered}{6} & \\ \hline &1&-2&2&\color{orangered}{21}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 21 } = \color{blue}{ 63 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-5&8&15&-2\\& & 3& -6& 6& \color{blue}{63} \\ \hline &1&-2&2&\color{blue}{21}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 63 } = \color{orangered}{ 61 } $
$$ \begin{array}{c|rrrrr}3&1&-5&8&15&\color{orangered}{ -2 }\\& & 3& -6& 6& \color{orangered}{63} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{2}&\color{blue}{21}&\color{orangered}{61} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-2x^{2}+2x+21 } $ with a remainder of $ \color{red}{ 61 } $.