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