Divide using synthetic division $$ ( x^{3}-2x^{2}-13x+13 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&1&-2&-13&13\\& & -3& 15& \color{black}{-6} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{2}&\color{orangered}{7} \end{array} $$The solution is:
$$ \frac{ x^{3}-2x^{2}-13x+13 }{ x+3 } = \color{blue}{x^{2}-5x+2} ~+~ \frac{ \color{red}{ 7 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&-2&-13&13\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 1 }&-2&-13&13\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&-2&-13&13\\& & \color{blue}{-3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}-3&1&\color{orangered}{ -2 }&-13&13\\& & \color{orangered}{-3} & & \\ \hline &1&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&-2&-13&13\\& & -3& \color{blue}{15} & \\ \hline &1&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 15 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}-3&1&-2&\color{orangered}{ -13 }&13\\& & -3& \color{orangered}{15} & \\ \hline &1&-5&\color{orangered}{2}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&-2&-13&13\\& & -3& 15& \color{blue}{-6} \\ \hline &1&-5&\color{blue}{2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}-3&1&-2&-13&\color{orangered}{ 13 }\\& & -3& 15& \color{orangered}{-6} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{2}&\color{orangered}{7} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-5x+2 } $ with a remainder of $ \color{red}{ 7 } $.