Divide using synthetic division $$ ( x^{3}+3x^{2}-28x-54 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&1&3&-28&-54\\& & -5& 10& \color{black}{90} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{-18}&\color{orangered}{36} \end{array} $$The solution is:
$$ \frac{ x^{3}+3x^{2}-28x-54 }{ x+5 } = \color{blue}{x^{2}-2x-18} ~+~ \frac{ \color{red}{ 36 } }{ x+5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 5 = 0 $ ( $ x = \color{blue}{ -5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&3&-28&-54\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 1 }&3&-28&-54\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 1 } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&3&-28&-54\\& & \color{blue}{-5} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}-5&1&\color{orangered}{ 3 }&-28&-54\\& & \color{orangered}{-5} & & \\ \hline &1&\color{orangered}{-2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&3&-28&-54\\& & -5& \color{blue}{10} & \\ \hline &1&\color{blue}{-2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -28 } + \color{orangered}{ 10 } = \color{orangered}{ -18 } $
$$ \begin{array}{c|rrrr}-5&1&3&\color{orangered}{ -28 }&-54\\& & -5& \color{orangered}{10} & \\ \hline &1&-2&\color{orangered}{-18}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -18 \right) } = \color{blue}{ 90 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&3&-28&-54\\& & -5& 10& \color{blue}{90} \\ \hline &1&-2&\color{blue}{-18}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -54 } + \color{orangered}{ 90 } = \color{orangered}{ 36 } $
$$ \begin{array}{c|rrrr}-5&1&3&-28&\color{orangered}{ -54 }\\& & -5& 10& \color{orangered}{90} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{-18}&\color{orangered}{36} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-2x-18 } $ with a remainder of $ \color{red}{ 36 } $.