Divide using synthetic division $$ ( -16x^{3}+22x^{2}+6x-20 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&-16&22&6&-20\\& & -16& 6& \color{black}{12} \\ \hline &\color{blue}{-16}&\color{blue}{6}&\color{blue}{12}&\color{orangered}{-8} \end{array} $$The solution is:
$$ \frac{ -16x^{3}+22x^{2}+6x-20 }{ x-1 } = \color{blue}{-16x^{2}+6x+12} \color{red}{~-~} \frac{ \color{red}{ 8 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&-16&22&6&-20\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ -16 }&22&6&-20\\& & & & \\ \hline &\color{orangered}{-16}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&-16&22&6&-20\\& & \color{blue}{-16} & & \\ \hline &\color{blue}{-16}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 22 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}1&-16&\color{orangered}{ 22 }&6&-20\\& & \color{orangered}{-16} & & \\ \hline &-16&\color{orangered}{6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 6 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&-16&22&6&-20\\& & -16& \color{blue}{6} & \\ \hline &-16&\color{blue}{6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 6 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrr}1&-16&22&\color{orangered}{ 6 }&-20\\& & -16& \color{orangered}{6} & \\ \hline &-16&6&\color{orangered}{12}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 12 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&-16&22&6&-20\\& & -16& 6& \color{blue}{12} \\ \hline &-16&6&\color{blue}{12}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 12 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}1&-16&22&6&\color{orangered}{ -20 }\\& & -16& 6& \color{orangered}{12} \\ \hline &\color{blue}{-16}&\color{blue}{6}&\color{blue}{12}&\color{orangered}{-8} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -16x^{2}+6x+12 } $ with a remainder of $ \color{red}{ -8 } $.