Divide using synthetic division $$ ( 4x^{3}-24x^{2}+21x-5 ) \div ( 2x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&4&-24&21&-5\\& & 4& -20& \color{black}{1} \\ \hline &\color{blue}{4}&\color{blue}{-20}&\color{blue}{1}&\color{orangered}{-4} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-24x^{2}+21x-5 }{ x-1 } = \color{blue}{4x^{2}-20x+1} \color{red}{~-~} \frac{ \color{red}{ 4 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&4&-24&21&-5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 4 }&-24&21&-5\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&4&-24&21&-5\\& & \color{blue}{4} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -24 } + \color{orangered}{ 4 } = \color{orangered}{ -20 } $
$$ \begin{array}{c|rrrr}1&4&\color{orangered}{ -24 }&21&-5\\& & \color{orangered}{4} & & \\ \hline &4&\color{orangered}{-20}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -20 \right) } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&4&-24&21&-5\\& & 4& \color{blue}{-20} & \\ \hline &4&\color{blue}{-20}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}1&4&-24&\color{orangered}{ 21 }&-5\\& & 4& \color{orangered}{-20} & \\ \hline &4&-20&\color{orangered}{1}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 1 } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&4&-24&21&-5\\& & 4& -20& \color{blue}{1} \\ \hline &4&-20&\color{blue}{1}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 1 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}1&4&-24&21&\color{orangered}{ -5 }\\& & 4& -20& \color{orangered}{1} \\ \hline &\color{blue}{4}&\color{blue}{-20}&\color{blue}{1}&\color{orangered}{-4} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-20x+1 } $ with a remainder of $ \color{red}{ -4 } $.