Divide using synthetic division $$ ( 40x^{3}-84x^{2}+56x-16 ) \div ( 4x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&40&-84&56&-16\\& & 80& -8& \color{black}{96} \\ \hline &\color{blue}{40}&\color{blue}{-4}&\color{blue}{48}&\color{orangered}{80} \end{array} $$The solution is:
$$ \frac{ 40x^{3}-84x^{2}+56x-16 }{ x-2 } = \color{blue}{40x^{2}-4x+48} ~+~ \frac{ \color{red}{ 80 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&40&-84&56&-16\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 40 }&-84&56&-16\\& & & & \\ \hline &\color{orangered}{40}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 40 } = \color{blue}{ 80 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&40&-84&56&-16\\& & \color{blue}{80} & & \\ \hline &\color{blue}{40}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -84 } + \color{orangered}{ 80 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}2&40&\color{orangered}{ -84 }&56&-16\\& & \color{orangered}{80} & & \\ \hline &40&\color{orangered}{-4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&40&-84&56&-16\\& & 80& \color{blue}{-8} & \\ \hline &40&\color{blue}{-4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 56 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 48 } $
$$ \begin{array}{c|rrrr}2&40&-84&\color{orangered}{ 56 }&-16\\& & 80& \color{orangered}{-8} & \\ \hline &40&-4&\color{orangered}{48}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 48 } = \color{blue}{ 96 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&40&-84&56&-16\\& & 80& -8& \color{blue}{96} \\ \hline &40&-4&\color{blue}{48}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 96 } = \color{orangered}{ 80 } $
$$ \begin{array}{c|rrrr}2&40&-84&56&\color{orangered}{ -16 }\\& & 80& -8& \color{orangered}{96} \\ \hline &\color{blue}{40}&\color{blue}{-4}&\color{blue}{48}&\color{orangered}{80} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 40x^{2}-4x+48 } $ with a remainder of $ \color{red}{ 80 } $.