Divide using synthetic division $$ ( 2x^{3}-12x^{2}+26x-20 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&2&-12&26&-20\\& & 4& -16& \color{black}{20} \\ \hline &\color{blue}{2}&\color{blue}{-8}&\color{blue}{10}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-12x^{2}+26x-20 }{ x-2 } = \color{blue}{2x^{2}-8x+10} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&2&-12&26&-20\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 2 }&-12&26&-20\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&2&-12&26&-20\\& & \color{blue}{4} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 4 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}2&2&\color{orangered}{ -12 }&26&-20\\& & \color{orangered}{4} & & \\ \hline &2&\color{orangered}{-8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&2&-12&26&-20\\& & 4& \color{blue}{-16} & \\ \hline &2&\color{blue}{-8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 26 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}2&2&-12&\color{orangered}{ 26 }&-20\\& & 4& \color{orangered}{-16} & \\ \hline &2&-8&\color{orangered}{10}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 10 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&2&-12&26&-20\\& & 4& -16& \color{blue}{20} \\ \hline &2&-8&\color{blue}{10}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 20 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}2&2&-12&26&\color{orangered}{ -20 }\\& & 4& -16& \color{orangered}{20} \\ \hline &\color{blue}{2}&\color{blue}{-8}&\color{blue}{10}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-8x+10 } $ with a remainder of $ \color{red}{ 0 } $.