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