Divide using synthetic division $$ ( 4x^{3}-27x^{2}+54x-31 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&4&-27&54&-31\\& & 4& -23& \color{black}{31} \\ \hline &\color{blue}{4}&\color{blue}{-23}&\color{blue}{31}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-27x^{2}+54x-31 }{ x-1 } = \color{blue}{4x^{2}-23x+31} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&4&-27&54&-31\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 4 }&-27&54&-31\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 4 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&4&-27&54&-31\\& & \color{blue}{4} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -27 } + \color{orangered}{ 4 } = \color{orangered}{ -23 } $
$$ \begin{array}{c|rrrr}1&4&\color{orangered}{ -27 }&54&-31\\& & \color{orangered}{4} & & \\ \hline &4&\color{orangered}{-23}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -23 \right) } = \color{blue}{ -23 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&4&-27&54&-31\\& & 4& \color{blue}{-23} & \\ \hline &4&\color{blue}{-23}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 54 } + \color{orangered}{ \left( -23 \right) } = \color{orangered}{ 31 } $
$$ \begin{array}{c|rrrr}1&4&-27&\color{orangered}{ 54 }&-31\\& & 4& \color{orangered}{-23} & \\ \hline &4&-23&\color{orangered}{31}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 31 } = \color{blue}{ 31 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&4&-27&54&-31\\& & 4& -23& \color{blue}{31} \\ \hline &4&-23&\color{blue}{31}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -31 } + \color{orangered}{ 31 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}1&4&-27&54&\color{orangered}{ -31 }\\& & 4& -23& \color{orangered}{31} \\ \hline &\color{blue}{4}&\color{blue}{-23}&\color{blue}{31}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-23x+31 } $ with a remainder of $ \color{red}{ 0 } $.