Divide using synthetic division $$ ( 2x^{3}+13x^{2}+17x-22 ) \div ( 2x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&2&13&17&-22\\& & 2& 15& \color{black}{32} \\ \hline &\color{blue}{2}&\color{blue}{15}&\color{blue}{32}&\color{orangered}{10} \end{array} $$The solution is:
$$ \frac{ 2x^{3}+13x^{2}+17x-22 }{ x-1 } = \color{blue}{2x^{2}+15x+32} ~+~ \frac{ \color{red}{ 10 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&13&17&-22\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 2 }&13&17&-22\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&13&17&-22\\& & \color{blue}{2} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ 2 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrr}1&2&\color{orangered}{ 13 }&17&-22\\& & \color{orangered}{2} & & \\ \hline &2&\color{orangered}{15}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 15 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&13&17&-22\\& & 2& \color{blue}{15} & \\ \hline &2&\color{blue}{15}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 17 } + \color{orangered}{ 15 } = \color{orangered}{ 32 } $
$$ \begin{array}{c|rrrr}1&2&13&\color{orangered}{ 17 }&-22\\& & 2& \color{orangered}{15} & \\ \hline &2&15&\color{orangered}{32}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 32 } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&13&17&-22\\& & 2& 15& \color{blue}{32} \\ \hline &2&15&\color{blue}{32}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -22 } + \color{orangered}{ 32 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}1&2&13&17&\color{orangered}{ -22 }\\& & 2& 15& \color{orangered}{32} \\ \hline &\color{blue}{2}&\color{blue}{15}&\color{blue}{32}&\color{orangered}{10} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}+15x+32 } $ with a remainder of $ \color{red}{ 10 } $.