Divide using synthetic division $$ ( 4x^{3}-x^{2}-15x+1 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&4&-1&-15&1\\& & 8& 14& \color{black}{-2} \\ \hline &\color{blue}{4}&\color{blue}{7}&\color{blue}{-1}&\color{orangered}{-1} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-x^{2}-15x+1 }{ x-2 } = \color{blue}{4x^{2}+7x-1} \color{red}{~-~} \frac{ \color{red}{ 1 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&4&-1&-15&1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 4 }&-1&-15&1\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 4 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&4&-1&-15&1\\& & \color{blue}{8} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 8 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}2&4&\color{orangered}{ -1 }&-15&1\\& & \color{orangered}{8} & & \\ \hline &4&\color{orangered}{7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 7 } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&4&-1&-15&1\\& & 8& \color{blue}{14} & \\ \hline &4&\color{blue}{7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 14 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}2&4&-1&\color{orangered}{ -15 }&1\\& & 8& \color{orangered}{14} & \\ \hline &4&7&\color{orangered}{-1}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&4&-1&-15&1\\& & 8& 14& \color{blue}{-2} \\ \hline &4&7&\color{blue}{-1}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}2&4&-1&-15&\color{orangered}{ 1 }\\& & 8& 14& \color{orangered}{-2} \\ \hline &\color{blue}{4}&\color{blue}{7}&\color{blue}{-1}&\color{orangered}{-1} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+7x-1 } $ with a remainder of $ \color{red}{ -1 } $.