Divide using synthetic division $$ ( 16x^{3}-20x^{2}-4x+15 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&16&-20&-4&15\\& & 16& -4& \color{black}{-8} \\ \hline &\color{blue}{16}&\color{blue}{-4}&\color{blue}{-8}&\color{orangered}{7} \end{array} $$The solution is:
$$ \frac{ 16x^{3}-20x^{2}-4x+15 }{ x-1 } = \color{blue}{16x^{2}-4x-8} ~+~ \frac{ \color{red}{ 7 } }{ x-1 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{1}&16&-20&-4&15\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 16 }&-20&-4&15\\& & & & \\ \hline &\color{orangered}{16}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 16 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&16&-20&-4&15\\& & \color{blue}{16} & & \\ \hline &\color{blue}{16}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 16 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}1&16&\color{orangered}{ -20 }&-4&15\\& & \color{orangered}{16} & & \\ \hline &16&\color{orangered}{-4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&16&-20&-4&15\\& & 16& \color{blue}{-4} & \\ \hline &16&\color{blue}{-4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}1&16&-20&\color{orangered}{ -4 }&15\\& & 16& \color{orangered}{-4} & \\ \hline &16&-4&\color{orangered}{-8}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&16&-20&-4&15\\& & 16& -4& \color{blue}{-8} \\ \hline &16&-4&\color{blue}{-8}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}1&16&-20&-4&\color{orangered}{ 15 }\\& & 16& -4& \color{orangered}{-8} \\ \hline &\color{blue}{16}&\color{blue}{-4}&\color{blue}{-8}&\color{orangered}{7} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 16x^{2}-4x-8 } $ with a remainder of $ \color{red}{ 7 } $.