Divide using synthetic division $$ ( 5x^{3}-9x^{2}-42x-8 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&5&-9&-42&-8\\& & 20& 44& \color{black}{8} \\ \hline &\color{blue}{5}&\color{blue}{11}&\color{blue}{2}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 5x^{3}-9x^{2}-42x-8 }{ x-4 } = \color{blue}{5x^{2}+11x+2} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{4}&5&-9&-42&-8\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 5 }&-9&-42&-8\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 5 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&5&-9&-42&-8\\& & \color{blue}{20} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 20 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrr}4&5&\color{orangered}{ -9 }&-42&-8\\& & \color{orangered}{20} & & \\ \hline &5&\color{orangered}{11}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 11 } = \color{blue}{ 44 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&5&-9&-42&-8\\& & 20& \color{blue}{44} & \\ \hline &5&\color{blue}{11}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -42 } + \color{orangered}{ 44 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}4&5&-9&\color{orangered}{ -42 }&-8\\& & 20& \color{orangered}{44} & \\ \hline &5&11&\color{orangered}{2}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 2 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&5&-9&-42&-8\\& & 20& 44& \color{blue}{8} \\ \hline &5&11&\color{blue}{2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 8 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}4&5&-9&-42&\color{orangered}{ -8 }\\& & 20& 44& \color{orangered}{8} \\ \hline &\color{blue}{5}&\color{blue}{11}&\color{blue}{2}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{2}+11x+2 } $ with a remainder of $ \color{red}{ 0 } $.