Divide using synthetic division $$ ( 8x^{3}+42x^{2}+15x+34 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&8&42&15&34\\& & -40& -10& \color{black}{-25} \\ \hline &\color{blue}{8}&\color{blue}{2}&\color{blue}{5}&\color{orangered}{9} \end{array} $$The solution is:
$$ \frac{ 8x^{3}+42x^{2}+15x+34 }{ x+5 } = \color{blue}{8x^{2}+2x+5} ~+~ \frac{ \color{red}{ 9 } }{ x+5 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&8&42&15&34\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 8 }&42&15&34\\& & & & \\ \hline &\color{orangered}{8}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 8 } = \color{blue}{ -40 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&8&42&15&34\\& & \color{blue}{-40} & & \\ \hline &\color{blue}{8}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 42 } + \color{orangered}{ \left( -40 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}-5&8&\color{orangered}{ 42 }&15&34\\& & \color{orangered}{-40} & & \\ \hline &8&\color{orangered}{2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 2 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&8&42&15&34\\& & -40& \color{blue}{-10} & \\ \hline &8&\color{blue}{2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}-5&8&42&\color{orangered}{ 15 }&34\\& & -40& \color{orangered}{-10} & \\ \hline &8&2&\color{orangered}{5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 5 } = \color{blue}{ -25 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&8&42&15&34\\& & -40& -10& \color{blue}{-25} \\ \hline &8&2&\color{blue}{5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 34 } + \color{orangered}{ \left( -25 \right) } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrr}-5&8&42&15&\color{orangered}{ 34 }\\& & -40& -10& \color{orangered}{-25} \\ \hline &\color{blue}{8}&\color{blue}{2}&\color{blue}{5}&\color{orangered}{9} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 8x^{2}+2x+5 } $ with a remainder of $ \color{red}{ 9 } $.