Divide using synthetic division $$ ( 16x^{3}-20x^{2}-4x+15 ) \div ( x-8 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}8&16&-20&-4&15\\& & 128& 864& \color{black}{6880} \\ \hline &\color{blue}{16}&\color{blue}{108}&\color{blue}{860}&\color{orangered}{6895} \end{array} $$The solution is:
$$ \frac{ 16x^{3}-20x^{2}-4x+15 }{ x-8 } = \color{blue}{16x^{2}+108x+860} ~+~ \frac{ \color{red}{ 6895 } }{ x-8 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{8}&16&-20&-4&15\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}8&\color{orangered}{ 16 }&-20&-4&15\\& & & & \\ \hline &\color{orangered}{16}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 16 } = \color{blue}{ 128 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&16&-20&-4&15\\& & \color{blue}{128} & & \\ \hline &\color{blue}{16}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 128 } = \color{orangered}{ 108 } $
$$ \begin{array}{c|rrrr}8&16&\color{orangered}{ -20 }&-4&15\\& & \color{orangered}{128} & & \\ \hline &16&\color{orangered}{108}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 108 } = \color{blue}{ 864 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&16&-20&-4&15\\& & 128& \color{blue}{864} & \\ \hline &16&\color{blue}{108}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 864 } = \color{orangered}{ 860 } $
$$ \begin{array}{c|rrrr}8&16&-20&\color{orangered}{ -4 }&15\\& & 128& \color{orangered}{864} & \\ \hline &16&108&\color{orangered}{860}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 860 } = \color{blue}{ 6880 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&16&-20&-4&15\\& & 128& 864& \color{blue}{6880} \\ \hline &16&108&\color{blue}{860}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ 6880 } = \color{orangered}{ 6895 } $
$$ \begin{array}{c|rrrr}8&16&-20&-4&\color{orangered}{ 15 }\\& & 128& 864& \color{orangered}{6880} \\ \hline &\color{blue}{16}&\color{blue}{108}&\color{blue}{860}&\color{orangered}{6895} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 16x^{2}+108x+860 } $ with a remainder of $ \color{red}{ 6895 } $.