Divide using synthetic division $$ ( 2x^{3}-14x^{2}+32x-20 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&2&-14&32&-20\\& & 2& -12& \color{black}{20} \\ \hline &\color{blue}{2}&\color{blue}{-12}&\color{blue}{20}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-14x^{2}+32x-20 }{ x-1 } = \color{blue}{2x^{2}-12x+20} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&-14&32&-20\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 2 }&-14&32&-20\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 2 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&-14&32&-20\\& & \color{blue}{2} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 2 } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrr}1&2&\color{orangered}{ -14 }&32&-20\\& & \color{orangered}{2} & & \\ \hline &2&\color{orangered}{-12}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&-14&32&-20\\& & 2& \color{blue}{-12} & \\ \hline &2&\color{blue}{-12}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 32 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrr}1&2&-14&\color{orangered}{ 32 }&-20\\& & 2& \color{orangered}{-12} & \\ \hline &2&-12&\color{orangered}{20}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 20 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&-14&32&-20\\& & 2& -12& \color{blue}{20} \\ \hline &2&-12&\color{blue}{20}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 20 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}1&2&-14&32&\color{orangered}{ -20 }\\& & 2& -12& \color{orangered}{20} \\ \hline &\color{blue}{2}&\color{blue}{-12}&\color{blue}{20}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-12x+20 } $ with a remainder of $ \color{red}{ 0 } $.