Divide using synthetic division $$ ( 4x^{4}-15x^{2}-4 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&4&0&-15&0&-4\\& & 12& 36& 63& \color{black}{189} \\ \hline &\color{blue}{4}&\color{blue}{12}&\color{blue}{21}&\color{blue}{63}&\color{orangered}{185} \end{array} $$The solution is:
$$ \frac{ 4x^{4}-15x^{2}-4 }{ x-3 } = \color{blue}{4x^{3}+12x^{2}+21x+63} ~+~ \frac{ \color{red}{ 185 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&0&-15&0&-4\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 4 }&0&-15&0&-4\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&0&-15&0&-4\\& & \color{blue}{12} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 12 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}3&4&\color{orangered}{ 0 }&-15&0&-4\\& & \color{orangered}{12} & & & \\ \hline &4&\color{orangered}{12}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 12 } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&0&-15&0&-4\\& & 12& \color{blue}{36} & & \\ \hline &4&\color{blue}{12}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 36 } = \color{orangered}{ 21 } $
$$ \begin{array}{c|rrrrr}3&4&0&\color{orangered}{ -15 }&0&-4\\& & 12& \color{orangered}{36} & & \\ \hline &4&12&\color{orangered}{21}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 21 } = \color{blue}{ 63 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&0&-15&0&-4\\& & 12& 36& \color{blue}{63} & \\ \hline &4&12&\color{blue}{21}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 63 } = \color{orangered}{ 63 } $
$$ \begin{array}{c|rrrrr}3&4&0&-15&\color{orangered}{ 0 }&-4\\& & 12& 36& \color{orangered}{63} & \\ \hline &4&12&21&\color{orangered}{63}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 63 } = \color{blue}{ 189 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&0&-15&0&-4\\& & 12& 36& 63& \color{blue}{189} \\ \hline &4&12&21&\color{blue}{63}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 189 } = \color{orangered}{ 185 } $
$$ \begin{array}{c|rrrrr}3&4&0&-15&0&\color{orangered}{ -4 }\\& & 12& 36& 63& \color{orangered}{189} \\ \hline &\color{blue}{4}&\color{blue}{12}&\color{blue}{21}&\color{blue}{63}&\color{orangered}{185} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}+12x^{2}+21x+63 } $ with a remainder of $ \color{red}{ 185 } $.