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