Divide using synthetic division $$ ( 15x^{3}-29x^{2}+6x+8 ) \div ( 5x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&15&-29&6&8\\& & -30& 118& \color{black}{-248} \\ \hline &\color{blue}{15}&\color{blue}{-59}&\color{blue}{124}&\color{orangered}{-240} \end{array} $$The solution is:
$$ \frac{ 15x^{3}-29x^{2}+6x+8 }{ x+2 } = \color{blue}{15x^{2}-59x+124} \color{red}{~-~} \frac{ \color{red}{ 240 } }{ x+2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&15&-29&6&8\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 15 }&-29&6&8\\& & & & \\ \hline &\color{orangered}{15}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 15 } = \color{blue}{ -30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&15&-29&6&8\\& & \color{blue}{-30} & & \\ \hline &\color{blue}{15}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -29 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ -59 } $
$$ \begin{array}{c|rrrr}-2&15&\color{orangered}{ -29 }&6&8\\& & \color{orangered}{-30} & & \\ \hline &15&\color{orangered}{-59}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -59 \right) } = \color{blue}{ 118 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&15&-29&6&8\\& & -30& \color{blue}{118} & \\ \hline &15&\color{blue}{-59}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 118 } = \color{orangered}{ 124 } $
$$ \begin{array}{c|rrrr}-2&15&-29&\color{orangered}{ 6 }&8\\& & -30& \color{orangered}{118} & \\ \hline &15&-59&\color{orangered}{124}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 124 } = \color{blue}{ -248 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&15&-29&6&8\\& & -30& 118& \color{blue}{-248} \\ \hline &15&-59&\color{blue}{124}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -248 \right) } = \color{orangered}{ -240 } $
$$ \begin{array}{c|rrrr}-2&15&-29&6&\color{orangered}{ 8 }\\& & -30& 118& \color{orangered}{-248} \\ \hline &\color{blue}{15}&\color{blue}{-59}&\color{blue}{124}&\color{orangered}{-240} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 15x^{2}-59x+124 } $ with a remainder of $ \color{red}{ -240 } $.