Divide using synthetic division $$ ( 16x^{3}-20x^{2}-4x+15 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&16&-20&-4&15\\& & 32& 24& \color{black}{40} \\ \hline &\color{blue}{16}&\color{blue}{12}&\color{blue}{20}&\color{orangered}{55} \end{array} $$The solution is:
$$ \frac{ 16x^{3}-20x^{2}-4x+15 }{ x-2 } = \color{blue}{16x^{2}+12x+20} ~+~ \frac{ \color{red}{ 55 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&16&-20&-4&15\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 16 }&-20&-4&15\\& & & & \\ \hline &\color{orangered}{16}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 16 } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&16&-20&-4&15\\& & \color{blue}{32} & & \\ \hline &\color{blue}{16}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 32 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrr}2&16&\color{orangered}{ -20 }&-4&15\\& & \color{orangered}{32} & & \\ \hline &16&\color{orangered}{12}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 12 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&16&-20&-4&15\\& & 32& \color{blue}{24} & \\ \hline &16&\color{blue}{12}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 24 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrr}2&16&-20&\color{orangered}{ -4 }&15\\& & 32& \color{orangered}{24} & \\ \hline &16&12&\color{orangered}{20}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 20 } = \color{blue}{ 40 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&16&-20&-4&15\\& & 32& 24& \color{blue}{40} \\ \hline &16&12&\color{blue}{20}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ 40 } = \color{orangered}{ 55 } $
$$ \begin{array}{c|rrrr}2&16&-20&-4&\color{orangered}{ 15 }\\& & 32& 24& \color{orangered}{40} \\ \hline &\color{blue}{16}&\color{blue}{12}&\color{blue}{20}&\color{orangered}{55} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 16x^{2}+12x+20 } $ with a remainder of $ \color{red}{ 55 } $.