Divide using synthetic division $$ ( 5x^{4}-8x^{2}+1 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&5&0&-8&0&1\\& & 10& 20& 24& \color{black}{48} \\ \hline &\color{blue}{5}&\color{blue}{10}&\color{blue}{12}&\color{blue}{24}&\color{orangered}{49} \end{array} $$The solution is:
$$ \frac{ 5x^{4}-8x^{2}+1 }{ x-2 } = \color{blue}{5x^{3}+10x^{2}+12x+24} ~+~ \frac{ \color{red}{ 49 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&0&-8&0&1\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 5 }&0&-8&0&1\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 5 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&0&-8&0&1\\& & \color{blue}{10} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 10 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}2&5&\color{orangered}{ 0 }&-8&0&1\\& & \color{orangered}{10} & & & \\ \hline &5&\color{orangered}{10}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 10 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&0&-8&0&1\\& & 10& \color{blue}{20} & & \\ \hline &5&\color{blue}{10}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 20 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}2&5&0&\color{orangered}{ -8 }&0&1\\& & 10& \color{orangered}{20} & & \\ \hline &5&10&\color{orangered}{12}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 12 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&0&-8&0&1\\& & 10& 20& \color{blue}{24} & \\ \hline &5&10&\color{blue}{12}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 24 } = \color{orangered}{ 24 } $
$$ \begin{array}{c|rrrrr}2&5&0&-8&\color{orangered}{ 0 }&1\\& & 10& 20& \color{orangered}{24} & \\ \hline &5&10&12&\color{orangered}{24}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 24 } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&0&-8&0&1\\& & 10& 20& 24& \color{blue}{48} \\ \hline &5&10&12&\color{blue}{24}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 48 } = \color{orangered}{ 49 } $
$$ \begin{array}{c|rrrrr}2&5&0&-8&0&\color{orangered}{ 1 }\\& & 10& 20& 24& \color{orangered}{48} \\ \hline &\color{blue}{5}&\color{blue}{10}&\color{blue}{12}&\color{blue}{24}&\color{orangered}{49} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{3}+10x^{2}+12x+24 } $ with a remainder of $ \color{red}{ 49 } $.