Divide using synthetic division $$ ( 4x^{3}-8x^{2}+2x-1 ) \div ( 2x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&4&-8&2&-1\\& & -4& 12& \color{black}{-14} \\ \hline &\color{blue}{4}&\color{blue}{-12}&\color{blue}{14}&\color{orangered}{-15} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-8x^{2}+2x-1 }{ x+1 } = \color{blue}{4x^{2}-12x+14} \color{red}{~-~} \frac{ \color{red}{ 15 } }{ x+1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&4&-8&2&-1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 4 }&-8&2&-1\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 4 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&4&-8&2&-1\\& & \color{blue}{-4} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrr}-1&4&\color{orangered}{ -8 }&2&-1\\& & \color{orangered}{-4} & & \\ \hline &4&\color{orangered}{-12}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&4&-8&2&-1\\& & -4& \color{blue}{12} & \\ \hline &4&\color{blue}{-12}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 12 } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrrr}-1&4&-8&\color{orangered}{ 2 }&-1\\& & -4& \color{orangered}{12} & \\ \hline &4&-12&\color{orangered}{14}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 14 } = \color{blue}{ -14 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&4&-8&2&-1\\& & -4& 12& \color{blue}{-14} \\ \hline &4&-12&\color{blue}{14}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -14 \right) } = \color{orangered}{ -15 } $
$$ \begin{array}{c|rrrr}-1&4&-8&2&\color{orangered}{ -1 }\\& & -4& 12& \color{orangered}{-14} \\ \hline &\color{blue}{4}&\color{blue}{-12}&\color{blue}{14}&\color{orangered}{-15} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-12x+14 } $ with a remainder of $ \color{red}{ -15 } $.