Divide using synthetic division $$ ( x^{3}-28x-48 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-4&1&0&-28&-48\\& & -4& 16& \color{black}{48} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{-12}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{3}-28x-48 }{ x+4 } = \color{blue}{x^{2}-4x-12} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&0&-28&-48\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-4&\color{orangered}{ 1 }&0&-28&-48\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&0&-28&-48\\& & \color{blue}{-4} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}-4&1&\color{orangered}{ 0 }&-28&-48\\& & \color{orangered}{-4} & & \\ \hline &1&\color{orangered}{-4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&0&-28&-48\\& & -4& \color{blue}{16} & \\ \hline &1&\color{blue}{-4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -28 } + \color{orangered}{ 16 } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrr}-4&1&0&\color{orangered}{ -28 }&-48\\& & -4& \color{orangered}{16} & \\ \hline &1&-4&\color{orangered}{-12}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&0&-28&-48\\& & -4& 16& \color{blue}{48} \\ \hline &1&-4&\color{blue}{-12}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -48 } + \color{orangered}{ 48 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-4&1&0&-28&\color{orangered}{ -48 }\\& & -4& 16& \color{orangered}{48} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{-12}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-4x-12 } $ with a remainder of $ \color{red}{ 0 } $.