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