Divide using synthetic division $$ ( 4x^{3}-6x^{2}-2x+11 ) \div ( 2x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&4&-6&-2&11\\& & -4& 10& \color{black}{-8} \\ \hline &\color{blue}{4}&\color{blue}{-10}&\color{blue}{8}&\color{orangered}{3} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-6x^{2}-2x+11 }{ x+1 } = \color{blue}{4x^{2}-10x+8} ~+~ \frac{ \color{red}{ 3 } }{ x+1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&4&-6&-2&11\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 4 }&-6&-2&11\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&4&-6&-2&11\\& & \color{blue}{-4} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrr}-1&4&\color{orangered}{ -6 }&-2&11\\& & \color{orangered}{-4} & & \\ \hline &4&\color{orangered}{-10}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&4&-6&-2&11\\& & -4& \color{blue}{10} & \\ \hline &4&\color{blue}{-10}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 10 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}-1&4&-6&\color{orangered}{ -2 }&11\\& & -4& \color{orangered}{10} & \\ \hline &4&-10&\color{orangered}{8}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 8 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&4&-6&-2&11\\& & -4& 10& \color{blue}{-8} \\ \hline &4&-10&\color{blue}{8}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}-1&4&-6&-2&\color{orangered}{ 11 }\\& & -4& 10& \color{orangered}{-8} \\ \hline &\color{blue}{4}&\color{blue}{-10}&\color{blue}{8}&\color{orangered}{3} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-10x+8 } $ with a remainder of $ \color{red}{ 3 } $.