Divide using synthetic division $$ ( x^{3}-12x^{2}+27x+40 ) \div ( x-8 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}8&1&-12&27&40\\& & 8& -32& \color{black}{-40} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{-5}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{3}-12x^{2}+27x+40 }{ x-8 } = \color{blue}{x^{2}-4x-5} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -8 = 0 $ ( $ x = \color{blue}{ 8 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{8}&1&-12&27&40\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}8&\color{orangered}{ 1 }&-12&27&40\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 1 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&1&-12&27&40\\& & \color{blue}{8} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 8 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}8&1&\color{orangered}{ -12 }&27&40\\& & \color{orangered}{8} & & \\ \hline &1&\color{orangered}{-4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -32 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&1&-12&27&40\\& & 8& \color{blue}{-32} & \\ \hline &1&\color{blue}{-4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 27 } + \color{orangered}{ \left( -32 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}8&1&-12&\color{orangered}{ 27 }&40\\& & 8& \color{orangered}{-32} & \\ \hline &1&-4&\color{orangered}{-5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -40 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&1&-12&27&40\\& & 8& -32& \color{blue}{-40} \\ \hline &1&-4&\color{blue}{-5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 40 } + \color{orangered}{ \left( -40 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}8&1&-12&27&\color{orangered}{ 40 }\\& & 8& -32& \color{orangered}{-40} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{-5}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-4x-5 } $ with a remainder of $ \color{red}{ 0 } $.