Divide using synthetic division $$ ( 5x^{3}+12x^{2}-42x-40 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-4&5&12&-42&-40\\& & -20& 32& \color{black}{40} \\ \hline &\color{blue}{5}&\color{blue}{-8}&\color{blue}{-10}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 5x^{3}+12x^{2}-42x-40 }{ x+4 } = \color{blue}{5x^{2}-8x-10} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&5&12&-42&-40\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-4&\color{orangered}{ 5 }&12&-42&-40\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&5&12&-42&-40\\& & \color{blue}{-20} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}-4&5&\color{orangered}{ 12 }&-42&-40\\& & \color{orangered}{-20} & & \\ \hline &5&\color{orangered}{-8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&5&12&-42&-40\\& & -20& \color{blue}{32} & \\ \hline &5&\color{blue}{-8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -42 } + \color{orangered}{ 32 } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrr}-4&5&12&\color{orangered}{ -42 }&-40\\& & -20& \color{orangered}{32} & \\ \hline &5&-8&\color{orangered}{-10}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 40 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&5&12&-42&-40\\& & -20& 32& \color{blue}{40} \\ \hline &5&-8&\color{blue}{-10}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -40 } + \color{orangered}{ 40 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-4&5&12&-42&\color{orangered}{ -40 }\\& & -20& 32& \color{orangered}{40} \\ \hline &\color{blue}{5}&\color{blue}{-8}&\color{blue}{-10}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{2}-8x-10 } $ with a remainder of $ \color{red}{ 0 } $.