Divide using synthetic division $$ ( 4x^{4}-15x^{2}-4 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&4&0&-15&0&-4\\& & 8& 16& 2& \color{black}{4} \\ \hline &\color{blue}{4}&\color{blue}{8}&\color{blue}{1}&\color{blue}{2}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 4x^{4}-15x^{2}-4 }{ x-2 } = \color{blue}{4x^{3}+8x^{2}+x+2} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&4&0&-15&0&-4\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 4 }&0&-15&0&-4\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 4 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&4&0&-15&0&-4\\& & \color{blue}{8} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 8 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}2&4&\color{orangered}{ 0 }&-15&0&-4\\& & \color{orangered}{8} & & & \\ \hline &4&\color{orangered}{8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 8 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&4&0&-15&0&-4\\& & 8& \color{blue}{16} & & \\ \hline &4&\color{blue}{8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 16 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}2&4&0&\color{orangered}{ -15 }&0&-4\\& & 8& \color{orangered}{16} & & \\ \hline &4&8&\color{orangered}{1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&4&0&-15&0&-4\\& & 8& 16& \color{blue}{2} & \\ \hline &4&8&\color{blue}{1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 2 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}2&4&0&-15&\color{orangered}{ 0 }&-4\\& & 8& 16& \color{orangered}{2} & \\ \hline &4&8&1&\color{orangered}{2}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&4&0&-15&0&-4\\& & 8& 16& 2& \color{blue}{4} \\ \hline &4&8&1&\color{blue}{2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 4 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}2&4&0&-15&0&\color{orangered}{ -4 }\\& & 8& 16& 2& \color{orangered}{4} \\ \hline &\color{blue}{4}&\color{blue}{8}&\color{blue}{1}&\color{blue}{2}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}+8x^{2}+x+2 } $ with a remainder of $ \color{red}{ 0 } $.