Divide using synthetic division $$ ( x^{6}-13x^{4}-52x^{2}+64 ) \div ( 1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrrr}0&1&0&-13&0&-52&0&64\\& & 0& 0& 0& 0& 0& \color{black}{0} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{-13}&\color{blue}{0}&\color{blue}{-52}&\color{blue}{0}&\color{orangered}{64} \end{array} $$The solution is:
$$ \frac{ x^{6}-13x^{4}-52x^{2}+64 }{ x } = \color{blue}{x^{5}-13x^{3}-52x} ~+~ \frac{ \color{red}{ 64 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrrrrr}\color{blue}{0}&1&0&-13&0&-52&0&64\\& & & & & & & \\ \hline &&&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrrr}0&\color{orangered}{ 1 }&0&-13&0&-52&0&64\\& & & & & & & \\ \hline &\color{orangered}{1}&&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 1 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{0}&1&0&-13&0&-52&0&64\\& & \color{blue}{0} & & & & & \\ \hline &\color{blue}{1}&&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrrr}0&1&\color{orangered}{ 0 }&-13&0&-52&0&64\\& & \color{orangered}{0} & & & & & \\ \hline &1&\color{orangered}{0}&&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{0}&1&0&-13&0&-52&0&64\\& & 0& \color{blue}{0} & & & & \\ \hline &1&\color{blue}{0}&&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 0 } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrrrrr}0&1&0&\color{orangered}{ -13 }&0&-52&0&64\\& & 0& \color{orangered}{0} & & & & \\ \hline &1&0&\color{orangered}{-13}&&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -13 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{0}&1&0&-13&0&-52&0&64\\& & 0& 0& \color{blue}{0} & & & \\ \hline &1&0&\color{blue}{-13}&&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrrr}0&1&0&-13&\color{orangered}{ 0 }&-52&0&64\\& & 0& 0& \color{orangered}{0} & & & \\ \hline &1&0&-13&\color{orangered}{0}&&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{0}&1&0&-13&0&-52&0&64\\& & 0& 0& 0& \color{blue}{0} & & \\ \hline &1&0&-13&\color{blue}{0}&&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -52 } + \color{orangered}{ 0 } = \color{orangered}{ -52 } $
$$ \begin{array}{c|rrrrrrr}0&1&0&-13&0&\color{orangered}{ -52 }&0&64\\& & 0& 0& 0& \color{orangered}{0} & & \\ \hline &1&0&-13&0&\color{orangered}{-52}&& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -52 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{0}&1&0&-13&0&-52&0&64\\& & 0& 0& 0& 0& \color{blue}{0} & \\ \hline &1&0&-13&0&\color{blue}{-52}&& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrrr}0&1&0&-13&0&-52&\color{orangered}{ 0 }&64\\& & 0& 0& 0& 0& \color{orangered}{0} & \\ \hline &1&0&-13&0&-52&\color{orangered}{0}& \end{array} $$Step 12 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{0}&1&0&-13&0&-52&0&64\\& & 0& 0& 0& 0& 0& \color{blue}{0} \\ \hline &1&0&-13&0&-52&\color{blue}{0}& \end{array} $$Step 13 : Add down last column: $ \color{orangered}{ 64 } + \color{orangered}{ 0 } = \color{orangered}{ 64 } $
$$ \begin{array}{c|rrrrrrr}0&1&0&-13&0&-52&0&\color{orangered}{ 64 }\\& & 0& 0& 0& 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{-13}&\color{blue}{0}&\color{blue}{-52}&\color{blue}{0}&\color{orangered}{64} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{5}-13x^{3}-52x } $ with a remainder of $ \color{red}{ 64 } $.