Divide using synthetic division $$ ( 2x^{4}+x^{3}-19x^{2}+5x+13 ) \div ( 2x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&2&1&-19&5&13\\& & 10& 55& 180& \color{black}{925} \\ \hline &\color{blue}{2}&\color{blue}{11}&\color{blue}{36}&\color{blue}{185}&\color{orangered}{938} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+x^{3}-19x^{2}+5x+13 }{ x-5 } = \color{blue}{2x^{3}+11x^{2}+36x+185} ~+~ \frac{ \color{red}{ 938 } }{ x-5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&2&1&-19&5&13\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 2 }&1&-19&5&13\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 2 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&2&1&-19&5&13\\& & \color{blue}{10} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 10 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrrr}5&2&\color{orangered}{ 1 }&-19&5&13\\& & \color{orangered}{10} & & & \\ \hline &2&\color{orangered}{11}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 11 } = \color{blue}{ 55 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&2&1&-19&5&13\\& & 10& \color{blue}{55} & & \\ \hline &2&\color{blue}{11}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -19 } + \color{orangered}{ 55 } = \color{orangered}{ 36 } $
$$ \begin{array}{c|rrrrr}5&2&1&\color{orangered}{ -19 }&5&13\\& & 10& \color{orangered}{55} & & \\ \hline &2&11&\color{orangered}{36}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 36 } = \color{blue}{ 180 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&2&1&-19&5&13\\& & 10& 55& \color{blue}{180} & \\ \hline &2&11&\color{blue}{36}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 180 } = \color{orangered}{ 185 } $
$$ \begin{array}{c|rrrrr}5&2&1&-19&\color{orangered}{ 5 }&13\\& & 10& 55& \color{orangered}{180} & \\ \hline &2&11&36&\color{orangered}{185}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 185 } = \color{blue}{ 925 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&2&1&-19&5&13\\& & 10& 55& 180& \color{blue}{925} \\ \hline &2&11&36&\color{blue}{185}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ 925 } = \color{orangered}{ 938 } $
$$ \begin{array}{c|rrrrr}5&2&1&-19&5&\color{orangered}{ 13 }\\& & 10& 55& 180& \color{orangered}{925} \\ \hline &\color{blue}{2}&\color{blue}{11}&\color{blue}{36}&\color{blue}{185}&\color{orangered}{938} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+11x^{2}+36x+185 } $ with a remainder of $ \color{red}{ 938 } $.