Divide using synthetic division $$ ( 7x^{3}+13x^{2}-23x ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&7&13&-23&0\\& & 7& 20& \color{black}{-3} \\ \hline &\color{blue}{7}&\color{blue}{20}&\color{blue}{-3}&\color{orangered}{-3} \end{array} $$The solution is:
$$ \frac{ 7x^{3}+13x^{2}-23x }{ x-1 } = \color{blue}{7x^{2}+20x-3} \color{red}{~-~} \frac{ \color{red}{ 3 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&7&13&-23&0\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 7 }&13&-23&0\\& & & & \\ \hline &\color{orangered}{7}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 7 } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&7&13&-23&0\\& & \color{blue}{7} & & \\ \hline &\color{blue}{7}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ 7 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrr}1&7&\color{orangered}{ 13 }&-23&0\\& & \color{orangered}{7} & & \\ \hline &7&\color{orangered}{20}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 20 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&7&13&-23&0\\& & 7& \color{blue}{20} & \\ \hline &7&\color{blue}{20}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -23 } + \color{orangered}{ 20 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}1&7&13&\color{orangered}{ -23 }&0\\& & 7& \color{orangered}{20} & \\ \hline &7&20&\color{orangered}{-3}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&7&13&-23&0\\& & 7& 20& \color{blue}{-3} \\ \hline &7&20&\color{blue}{-3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}1&7&13&-23&\color{orangered}{ 0 }\\& & 7& 20& \color{orangered}{-3} \\ \hline &\color{blue}{7}&\color{blue}{20}&\color{blue}{-3}&\color{orangered}{-3} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{2}+20x-3 } $ with a remainder of $ \color{red}{ -3 } $.