Divide using synthetic division $$ ( 2x^{5}-27x^{4}+167x^{3}-451x^{2}+210x-25 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}5&2&-27&167&-451&210&-25\\& & 10& -85& 410& -205& \color{black}{25} \\ \hline &\color{blue}{2}&\color{blue}{-17}&\color{blue}{82}&\color{blue}{-41}&\color{blue}{5}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{5}-27x^{4}+167x^{3}-451x^{2}+210x-25 }{ x-5 } = \color{blue}{2x^{4}-17x^{3}+82x^{2}-41x+5} $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{5}&2&-27&167&-451&210&-25\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}5&\color{orangered}{ 2 }&-27&167&-451&210&-25\\& & & & & & \\ \hline &\color{orangered}{2}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{5}&2&-27&167&-451&210&-25\\& & \color{blue}{10} & & & & \\ \hline &\color{blue}{2}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -27 } + \color{orangered}{ 10 } = \color{orangered}{ -17 } $
$$ \begin{array}{c|rrrrrr}5&2&\color{orangered}{ -27 }&167&-451&210&-25\\& & \color{orangered}{10} & & & & \\ \hline &2&\color{orangered}{-17}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -17 \right) } = \color{blue}{ -85 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{5}&2&-27&167&-451&210&-25\\& & 10& \color{blue}{-85} & & & \\ \hline &2&\color{blue}{-17}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 167 } + \color{orangered}{ \left( -85 \right) } = \color{orangered}{ 82 } $
$$ \begin{array}{c|rrrrrr}5&2&-27&\color{orangered}{ 167 }&-451&210&-25\\& & 10& \color{orangered}{-85} & & & \\ \hline &2&-17&\color{orangered}{82}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 82 } = \color{blue}{ 410 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{5}&2&-27&167&-451&210&-25\\& & 10& -85& \color{blue}{410} & & \\ \hline &2&-17&\color{blue}{82}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -451 } + \color{orangered}{ 410 } = \color{orangered}{ -41 } $
$$ \begin{array}{c|rrrrrr}5&2&-27&167&\color{orangered}{ -451 }&210&-25\\& & 10& -85& \color{orangered}{410} & & \\ \hline &2&-17&82&\color{orangered}{-41}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -41 \right) } = \color{blue}{ -205 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{5}&2&-27&167&-451&210&-25\\& & 10& -85& 410& \color{blue}{-205} & \\ \hline &2&-17&82&\color{blue}{-41}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 210 } + \color{orangered}{ \left( -205 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrrr}5&2&-27&167&-451&\color{orangered}{ 210 }&-25\\& & 10& -85& 410& \color{orangered}{-205} & \\ \hline &2&-17&82&-41&\color{orangered}{5}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 5 } = \color{blue}{ 25 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{5}&2&-27&167&-451&210&-25\\& & 10& -85& 410& -205& \color{blue}{25} \\ \hline &2&-17&82&-41&\color{blue}{5}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -25 } + \color{orangered}{ 25 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}5&2&-27&167&-451&210&\color{orangered}{ -25 }\\& & 10& -85& 410& -205& \color{orangered}{25} \\ \hline &\color{blue}{2}&\color{blue}{-17}&\color{blue}{82}&\color{blue}{-41}&\color{blue}{5}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{4}-17x^{3}+82x^{2}-41x+5 } $ with a remainder of $ \color{red}{ 0 } $.