Divide using synthetic division $$ ( 4x^{4}+11x^{3}-19x^{2}-4x+3 ) \div ( 4x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&4&11&-19&-4&3\\& & 4& 15& -4& \color{black}{-8} \\ \hline &\color{blue}{4}&\color{blue}{15}&\color{blue}{-4}&\color{blue}{-8}&\color{orangered}{-5} \end{array} $$The solution is:
$$ \frac{ 4x^{4}+11x^{3}-19x^{2}-4x+3 }{ x-1 } = \color{blue}{4x^{3}+15x^{2}-4x-8} \color{red}{~-~} \frac{ \color{red}{ 5 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&11&-19&-4&3\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 4 }&11&-19&-4&3\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 4 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&11&-19&-4&3\\& & \color{blue}{4} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ 4 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrrr}1&4&\color{orangered}{ 11 }&-19&-4&3\\& & \color{orangered}{4} & & & \\ \hline &4&\color{orangered}{15}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&11&-19&-4&3\\& & 4& \color{blue}{15} & & \\ \hline &4&\color{blue}{15}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -19 } + \color{orangered}{ 15 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}1&4&11&\color{orangered}{ -19 }&-4&3\\& & 4& \color{orangered}{15} & & \\ \hline &4&15&\color{orangered}{-4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&11&-19&-4&3\\& & 4& 15& \color{blue}{-4} & \\ \hline &4&15&\color{blue}{-4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}1&4&11&-19&\color{orangered}{ -4 }&3\\& & 4& 15& \color{orangered}{-4} & \\ \hline &4&15&-4&\color{orangered}{-8}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&11&-19&-4&3\\& & 4& 15& -4& \color{blue}{-8} \\ \hline &4&15&-4&\color{blue}{-8}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}1&4&11&-19&-4&\color{orangered}{ 3 }\\& & 4& 15& -4& \color{orangered}{-8} \\ \hline &\color{blue}{4}&\color{blue}{15}&\color{blue}{-4}&\color{blue}{-8}&\color{orangered}{-5} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}+15x^{2}-4x-8 } $ with a remainder of $ \color{red}{ -5 } $.