The synthetic division table is:
$$ \begin{array}{c|rrrr}0&2&-19&54&-45\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{2}&\color{blue}{-19}&\color{blue}{54}&\color{orangered}{-45} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-19x^{2}+54x-45 }{ x } = \color{blue}{2x^{2}-19x+54} \color{red}{~-~} \frac{ \color{red}{ 45 } }{ x } $$Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrr}\color{blue}{0}&2&-19&54&-45\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}0&\color{orangered}{ 2 }&-19&54&-45\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 2 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&2&-19&54&-45\\& & \color{blue}{0} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -19 } + \color{orangered}{ 0 } = \color{orangered}{ -19 } $
$$ \begin{array}{c|rrrr}0&2&\color{orangered}{ -19 }&54&-45\\& & \color{orangered}{0} & & \\ \hline &2&\color{orangered}{-19}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -19 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&2&-19&54&-45\\& & 0& \color{blue}{0} & \\ \hline &2&\color{blue}{-19}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 54 } + \color{orangered}{ 0 } = \color{orangered}{ 54 } $
$$ \begin{array}{c|rrrr}0&2&-19&\color{orangered}{ 54 }&-45\\& & 0& \color{orangered}{0} & \\ \hline &2&-19&\color{orangered}{54}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 54 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&2&-19&54&-45\\& & 0& 0& \color{blue}{0} \\ \hline &2&-19&\color{blue}{54}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -45 } + \color{orangered}{ 0 } = \color{orangered}{ -45 } $
$$ \begin{array}{c|rrrr}0&2&-19&54&\color{orangered}{ -45 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{2}&\color{blue}{-19}&\color{blue}{54}&\color{orangered}{-45} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-19x+54 } $ with a remainder of $ \color{red}{ -45 } $.