Divide using synthetic division $$ ( 16x^{2}-36x+13 ) \div ( 4x-1 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrr}1&16&-36&13\\& & 16& \color{black}{-20} \\ \hline &\color{blue}{16}&\color{blue}{-20}&\color{orangered}{-7} \end{array} $$The solution is:
$$ \dfrac{ 16x^{2}-36x+13 }{ x-1 } = \color{blue}{16x-20} \color{red}{~-~} \dfrac{ \color{red}{ 7 } }{ x-1 } $$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.
$$ \begin{array}{c|rrr}\color{blue}{1}&16&-36&13\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}1&\color{orangered}{ 16 }&-36&13\\& & & \\ \hline &\color{orangered}{16}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 16 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrr}\color{blue}{1}&16&-36&13\\& & \color{blue}{16} & \\ \hline &\color{blue}{16}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -36 } + \color{orangered}{ 16 } = \color{orangered}{ -20 } $
$$ \begin{array}{c|rrr}1&16&\color{orangered}{ -36 }&13\\& & \color{orangered}{16} & \\ \hline &16&\color{orangered}{-20}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -20 \right) } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrr}\color{blue}{1}&16&-36&13\\& & 16& \color{blue}{-20} \\ \hline &16&\color{blue}{-20}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrr}1&16&-36&\color{orangered}{ 13 }\\& & 16& \color{orangered}{-20} \\ \hline &\color{blue}{16}&\color{blue}{-20}&\color{orangered}{-7} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 16x-20 } $ with a remainder of $ \color{red}{ -7 } $.