Divide using synthetic division $$ ( 4x^{3}-7x-3 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&4&0&-7&-3\\& & 4& 4& \color{black}{-3} \\ \hline &\color{blue}{4}&\color{blue}{4}&\color{blue}{-3}&\color{orangered}{-6} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-7x-3 }{ x-1 } = \color{blue}{4x^{2}+4x-3} \color{red}{~-~} \frac{ \color{red}{ 6 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&4&0&-7&-3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 4 }&0&-7&-3\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&4&0&-7&-3\\& & \color{blue}{4} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 4 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}1&4&\color{orangered}{ 0 }&-7&-3\\& & \color{orangered}{4} & & \\ \hline &4&\color{orangered}{4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 4 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&4&0&-7&-3\\& & 4& \color{blue}{4} & \\ \hline &4&\color{blue}{4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 4 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}1&4&0&\color{orangered}{ -7 }&-3\\& & 4& \color{orangered}{4} & \\ \hline &4&4&\color{orangered}{-3}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&4&0&-7&-3\\& & 4& 4& \color{blue}{-3} \\ \hline &4&4&\color{blue}{-3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}1&4&0&-7&\color{orangered}{ -3 }\\& & 4& 4& \color{orangered}{-3} \\ \hline &\color{blue}{4}&\color{blue}{4}&\color{blue}{-3}&\color{orangered}{-6} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+4x-3 } $ with a remainder of $ \color{red}{ -6 } $.