Divide using synthetic division $$ ( 4x^{2}+23x-35 ) \div ( 4x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrr}5&4&23&-35\\& & 20& \color{black}{215} \\ \hline &\color{blue}{4}&\color{blue}{43}&\color{orangered}{180} \end{array} $$The solution is:
$$ \frac{ 4x^{2}+23x-35 }{ x-5 } = \color{blue}{4x+43} ~+~ \frac{ \color{red}{ 180 } }{ x-5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrr}\color{blue}{5}&4&23&-35\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}5&\color{orangered}{ 4 }&23&-35\\& & & \\ \hline &\color{orangered}{4}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 4 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrr}\color{blue}{5}&4&23&-35\\& & \color{blue}{20} & \\ \hline &\color{blue}{4}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 23 } + \color{orangered}{ 20 } = \color{orangered}{ 43 } $
$$ \begin{array}{c|rrr}5&4&\color{orangered}{ 23 }&-35\\& & \color{orangered}{20} & \\ \hline &4&\color{orangered}{43}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 43 } = \color{blue}{ 215 } $.
$$ \begin{array}{c|rrr}\color{blue}{5}&4&23&-35\\& & 20& \color{blue}{215} \\ \hline &4&\color{blue}{43}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -35 } + \color{orangered}{ 215 } = \color{orangered}{ 180 } $
$$ \begin{array}{c|rrr}5&4&23&\color{orangered}{ -35 }\\& & 20& \color{orangered}{215} \\ \hline &\color{blue}{4}&\color{blue}{43}&\color{orangered}{180} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x+43 } $ with a remainder of $ \color{red}{ 180 } $.