Divide using synthetic division $$ ( 2x^{4}-4x^{3}-14x+26 ) \div ( 2x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&2&-4&0&-14&26\\& & 8& 16& 64& \color{black}{200} \\ \hline &\color{blue}{2}&\color{blue}{4}&\color{blue}{16}&\color{blue}{50}&\color{orangered}{226} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-4x^{3}-14x+26 }{ x-4 } = \color{blue}{2x^{3}+4x^{2}+16x+50} ~+~ \frac{ \color{red}{ 226 } }{ x-4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-4&0&-14&26\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}4&\color{orangered}{ 2 }&-4&0&-14&26\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 2 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-4&0&-14&26\\& & \color{blue}{8} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 8 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}4&2&\color{orangered}{ -4 }&0&-14&26\\& & \color{orangered}{8} & & & \\ \hline &2&\color{orangered}{4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 4 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-4&0&-14&26\\& & 8& \color{blue}{16} & & \\ \hline &2&\color{blue}{4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 16 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrrr}4&2&-4&\color{orangered}{ 0 }&-14&26\\& & 8& \color{orangered}{16} & & \\ \hline &2&4&\color{orangered}{16}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 16 } = \color{blue}{ 64 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-4&0&-14&26\\& & 8& 16& \color{blue}{64} & \\ \hline &2&4&\color{blue}{16}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 64 } = \color{orangered}{ 50 } $
$$ \begin{array}{c|rrrrr}4&2&-4&0&\color{orangered}{ -14 }&26\\& & 8& 16& \color{orangered}{64} & \\ \hline &2&4&16&\color{orangered}{50}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 50 } = \color{blue}{ 200 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-4&0&-14&26\\& & 8& 16& 64& \color{blue}{200} \\ \hline &2&4&16&\color{blue}{50}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 26 } + \color{orangered}{ 200 } = \color{orangered}{ 226 } $
$$ \begin{array}{c|rrrrr}4&2&-4&0&-14&\color{orangered}{ 26 }\\& & 8& 16& 64& \color{orangered}{200} \\ \hline &\color{blue}{2}&\color{blue}{4}&\color{blue}{16}&\color{blue}{50}&\color{orangered}{226} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+4x^{2}+16x+50 } $ with a remainder of $ \color{red}{ 226 } $.