Divide using synthetic division $$ ( x^{3}-3x^{2}-15x+52 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-4&1&-3&-15&52\\& & -4& 28& \color{black}{-52} \\ \hline &\color{blue}{1}&\color{blue}{-7}&\color{blue}{13}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{3}-3x^{2}-15x+52 }{ x+4 } = \color{blue}{x^{2}-7x+13} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 4 = 0 $ ( $ x = \color{blue}{ -4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&-3&-15&52\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-4&\color{orangered}{ 1 }&-3&-15&52\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 1 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&-3&-15&52\\& & \color{blue}{-4} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}-4&1&\color{orangered}{ -3 }&-15&52\\& & \color{orangered}{-4} & & \\ \hline &1&\color{orangered}{-7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 28 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&-3&-15&52\\& & -4& \color{blue}{28} & \\ \hline &1&\color{blue}{-7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 28 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrr}-4&1&-3&\color{orangered}{ -15 }&52\\& & -4& \color{orangered}{28} & \\ \hline &1&-7&\color{orangered}{13}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 13 } = \color{blue}{ -52 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&-3&-15&52\\& & -4& 28& \color{blue}{-52} \\ \hline &1&-7&\color{blue}{13}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 52 } + \color{orangered}{ \left( -52 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-4&1&-3&-15&\color{orangered}{ 52 }\\& & -4& 28& \color{orangered}{-52} \\ \hline &\color{blue}{1}&\color{blue}{-7}&\color{blue}{13}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-7x+13 } $ with a remainder of $ \color{red}{ 0 } $.