Divide using synthetic division $$ ( 4x^{3}-9x^{2}-4x-15 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&4&-9&-4&-15\\& & 12& 9& \color{black}{15} \\ \hline &\color{blue}{4}&\color{blue}{3}&\color{blue}{5}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-9x^{2}-4x-15 }{ x-3 } = \color{blue}{4x^{2}+3x+5} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-9&-4&-15\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 4 }&-9&-4&-15\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-9&-4&-15\\& & \color{blue}{12} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 12 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}3&4&\color{orangered}{ -9 }&-4&-15\\& & \color{orangered}{12} & & \\ \hline &4&\color{orangered}{3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-9&-4&-15\\& & 12& \color{blue}{9} & \\ \hline &4&\color{blue}{3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 9 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}3&4&-9&\color{orangered}{ -4 }&-15\\& & 12& \color{orangered}{9} & \\ \hline &4&3&\color{orangered}{5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 5 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-9&-4&-15\\& & 12& 9& \color{blue}{15} \\ \hline &4&3&\color{blue}{5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 15 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}3&4&-9&-4&\color{orangered}{ -15 }\\& & 12& 9& \color{orangered}{15} \\ \hline &\color{blue}{4}&\color{blue}{3}&\color{blue}{5}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+3x+5 } $ with a remainder of $ \color{red}{ 0 } $.