Divide using synthetic division $$ ( 2x^{3}-19x^{2}+54x-45 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&2&-19&54&-45\\& & 2& -17& \color{black}{37} \\ \hline &\color{blue}{2}&\color{blue}{-17}&\color{blue}{37}&\color{orangered}{-8} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-19x^{2}+54x-45 }{ x-1 } = \color{blue}{2x^{2}-17x+37} \color{red}{~-~} \frac{ \color{red}{ 8 } }{ x-1 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&-19&54&-45\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 2 }&-19&54&-45\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 2 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&-19&54&-45\\& & \color{blue}{2} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -19 } + \color{orangered}{ 2 } = \color{orangered}{ -17 } $
$$ \begin{array}{c|rrrr}1&2&\color{orangered}{ -19 }&54&-45\\& & \color{orangered}{2} & & \\ \hline &2&\color{orangered}{-17}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -17 \right) } = \color{blue}{ -17 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&-19&54&-45\\& & 2& \color{blue}{-17} & \\ \hline &2&\color{blue}{-17}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 54 } + \color{orangered}{ \left( -17 \right) } = \color{orangered}{ 37 } $
$$ \begin{array}{c|rrrr}1&2&-19&\color{orangered}{ 54 }&-45\\& & 2& \color{orangered}{-17} & \\ \hline &2&-17&\color{orangered}{37}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 37 } = \color{blue}{ 37 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&-19&54&-45\\& & 2& -17& \color{blue}{37} \\ \hline &2&-17&\color{blue}{37}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -45 } + \color{orangered}{ 37 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}1&2&-19&54&\color{orangered}{ -45 }\\& & 2& -17& \color{orangered}{37} \\ \hline &\color{blue}{2}&\color{blue}{-17}&\color{blue}{37}&\color{orangered}{-8} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-17x+37 } $ with a remainder of $ \color{red}{ -8 } $.