Divide using synthetic division $$ ( 4x^{4}-5x^{2}+1 ) \div ( x-0.2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}\frac{ 1 }{ 5 }&4&0&-5&0&1\\& & \frac{ 4 }{ 5 }& \frac{ 4 }{ 25 }& -\frac{ 121 }{ 125 }& \color{black}{-\frac{ 121 }{ 625 }} \\ \hline &\color{blue}{4}&\color{blue}{\frac{ 4 }{ 5 }}&\color{blue}{-\frac{ 121 }{ 25 }}&\color{blue}{-\frac{ 121 }{ 125 }}&\color{orangered}{\frac{ 504 }{ 625 }} \end{array} $$The solution is:
$$ \frac{ 4x^{4}-5x^{2}+1 }{ x-\frac{ 1 }{ 5 } } = \color{blue}{4x^{3}+\frac{ 4 }{ 5 }x^{2}-\frac{ 121 }{ 25 }x-\frac{ 121 }{ 125 }} ~+~ \frac{ \color{red}{ \frac{ 504 }{ 625 } } }{ x-\frac{ 1 }{ 5 } } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -\frac{ 1 }{ 5 } = 0 $ ( $ x = \color{blue}{ \frac{ 1 }{ 5 } } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{\frac{ 1 }{ 5 }}&4&0&-5&0&1\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}\frac{ 1 }{ 5 }&\color{orangered}{ 4 }&0&-5&0&1\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ \frac{ 1 }{ 5 } } \cdot \color{blue}{ 4 } = \color{blue}{ \frac{ 4 }{ 5 } } $.
$$ \begin{array}{c|rrrrr}\color{blue}{\frac{ 1 }{ 5 }}&4&0&-5&0&1\\& & \color{blue}{\frac{ 4 }{ 5 }} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \frac{ 4 }{ 5 } } = \color{orangered}{ \frac{ 4 }{ 5 } } $
$$ \begin{array}{c|rrrrr}\frac{ 1 }{ 5 }&4&\color{orangered}{ 0 }&-5&0&1\\& & \color{orangered}{\frac{ 4 }{ 5 }} & & & \\ \hline &4&\color{orangered}{\frac{ 4 }{ 5 }}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ \frac{ 1 }{ 5 } } \cdot \color{blue}{ \frac{ 4 }{ 5 } } = \color{blue}{ \frac{ 4 }{ 25 } } $.
$$ \begin{array}{c|rrrrr}\color{blue}{\frac{ 1 }{ 5 }}&4&0&-5&0&1\\& & \frac{ 4 }{ 5 }& \color{blue}{\frac{ 4 }{ 25 }} & & \\ \hline &4&\color{blue}{\frac{ 4 }{ 5 }}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \frac{ 4 }{ 25 } } = \color{orangered}{ -\frac{ 121 }{ 25 } } $
$$ \begin{array}{c|rrrrr}\frac{ 1 }{ 5 }&4&0&\color{orangered}{ -5 }&0&1\\& & \frac{ 4 }{ 5 }& \color{orangered}{\frac{ 4 }{ 25 }} & & \\ \hline &4&\frac{ 4 }{ 5 }&\color{orangered}{-\frac{ 121 }{ 25 }}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ \frac{ 1 }{ 5 } } \cdot \color{blue}{ \left( -\frac{ 121 }{ 25 } \right) } = \color{blue}{ -\frac{ 121 }{ 125 } } $.
$$ \begin{array}{c|rrrrr}\color{blue}{\frac{ 1 }{ 5 }}&4&0&-5&0&1\\& & \frac{ 4 }{ 5 }& \frac{ 4 }{ 25 }& \color{blue}{-\frac{ 121 }{ 125 }} & \\ \hline &4&\frac{ 4 }{ 5 }&\color{blue}{-\frac{ 121 }{ 25 }}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -\frac{ 121 }{ 125 } \right) } = \color{orangered}{ -\frac{ 121 }{ 125 } } $
$$ \begin{array}{c|rrrrr}\frac{ 1 }{ 5 }&4&0&-5&\color{orangered}{ 0 }&1\\& & \frac{ 4 }{ 5 }& \frac{ 4 }{ 25 }& \color{orangered}{-\frac{ 121 }{ 125 }} & \\ \hline &4&\frac{ 4 }{ 5 }&-\frac{ 121 }{ 25 }&\color{orangered}{-\frac{ 121 }{ 125 }}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ \frac{ 1 }{ 5 } } \cdot \color{blue}{ \left( -\frac{ 121 }{ 125 } \right) } = \color{blue}{ -\frac{ 121 }{ 625 } } $.
$$ \begin{array}{c|rrrrr}\color{blue}{\frac{ 1 }{ 5 }}&4&0&-5&0&1\\& & \frac{ 4 }{ 5 }& \frac{ 4 }{ 25 }& -\frac{ 121 }{ 125 }& \color{blue}{-\frac{ 121 }{ 625 }} \\ \hline &4&\frac{ 4 }{ 5 }&-\frac{ 121 }{ 25 }&\color{blue}{-\frac{ 121 }{ 125 }}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -\frac{ 121 }{ 625 } \right) } = \color{orangered}{ \frac{ 504 }{ 625 } } $
$$ \begin{array}{c|rrrrr}\frac{ 1 }{ 5 }&4&0&-5&0&\color{orangered}{ 1 }\\& & \frac{ 4 }{ 5 }& \frac{ 4 }{ 25 }& -\frac{ 121 }{ 125 }& \color{orangered}{-\frac{ 121 }{ 625 }} \\ \hline &\color{blue}{4}&\color{blue}{\frac{ 4 }{ 5 }}&\color{blue}{-\frac{ 121 }{ 25 }}&\color{blue}{-\frac{ 121 }{ 125 }}&\color{orangered}{\frac{ 504 }{ 625 }} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}+\frac{ 4 }{ 5 }x^{2}-\frac{ 121 }{ 25 }x-\frac{ 121 }{ 125 } } $ with a remainder of $ \color{red}{ \frac{ 504 }{ 625 } } $.