Divide using synthetic division $$ ( 4x^{3}-13x^{2}-3 ) \div ( 4x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&4&-13&0&-3\\& & 12& -3& \color{black}{-9} \\ \hline &\color{blue}{4}&\color{blue}{-1}&\color{blue}{-3}&\color{orangered}{-12} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-13x^{2}-3 }{ x-3 } = \color{blue}{4x^{2}-x-3} \color{red}{~-~} \frac{ \color{red}{ 12 } }{ x-3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-13&0&-3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 4 }&-13&0&-3\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-13&0&-3\\& & \color{blue}{12} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 12 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}3&4&\color{orangered}{ -13 }&0&-3\\& & \color{orangered}{12} & & \\ \hline &4&\color{orangered}{-1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-13&0&-3\\& & 12& \color{blue}{-3} & \\ \hline &4&\color{blue}{-1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}3&4&-13&\color{orangered}{ 0 }&-3\\& & 12& \color{orangered}{-3} & \\ \hline &4&-1&\color{orangered}{-3}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-13&0&-3\\& & 12& -3& \color{blue}{-9} \\ \hline &4&-1&\color{blue}{-3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrr}3&4&-13&0&\color{orangered}{ -3 }\\& & 12& -3& \color{orangered}{-9} \\ \hline &\color{blue}{4}&\color{blue}{-1}&\color{blue}{-3}&\color{orangered}{-12} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-x-3 } $ with a remainder of $ \color{red}{ -12 } $.