Check if $ x-4 $ is a factor of $ 2x^{4}-15x^{3}+27x^{2}+2x+8 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&2&-15&27&2&8\\& & 8& -28& -4& \color{black}{-8} \\ \hline &\color{blue}{2}&\color{blue}{-7}&\color{blue}{-1}&\color{blue}{-2}&\color{orangered}{0} \end{array} $$Because the remainder equals zero, we conclude that the $ x-4 $ is a factor of the $ 2x^{4}-15x^{3}+27x^{2}+2x+8 $.
Explanation
First we need to create a synthetic division table.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-15&27&2&8\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}4&\color{orangered}{ 2 }&-15&27&2&8\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-15&27&2&8\\& & \color{blue}{8} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 8 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}4&2&\color{orangered}{ -15 }&27&2&8\\& & \color{orangered}{8} & & & \\ \hline &2&\color{orangered}{-7}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -28 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-15&27&2&8\\& & 8& \color{blue}{-28} & & \\ \hline &2&\color{blue}{-7}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 27 } + \color{orangered}{ \left( -28 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}4&2&-15&\color{orangered}{ 27 }&2&8\\& & 8& \color{orangered}{-28} & & \\ \hline &2&-7&\color{orangered}{-1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-15&27&2&8\\& & 8& -28& \color{blue}{-4} & \\ \hline &2&-7&\color{blue}{-1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}4&2&-15&27&\color{orangered}{ 2 }&8\\& & 8& -28& \color{orangered}{-4} & \\ \hline &2&-7&-1&\color{orangered}{-2}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-15&27&2&8\\& & 8& -28& -4& \color{blue}{-8} \\ \hline &2&-7&-1&\color{blue}{-2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}4&2&-15&27&2&\color{orangered}{ 8 }\\& & 8& -28& -4& \color{orangered}{-8} \\ \hline &\color{blue}{2}&\color{blue}{-7}&\color{blue}{-1}&\color{blue}{-2}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right)$.