Divide using synthetic division $$ ( -5x^{4}+3x^{3}+x^{2}-5 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&-5&3&1&0&-5\\& & -10& -14& -26& \color{black}{-52} \\ \hline &\color{blue}{-5}&\color{blue}{-7}&\color{blue}{-13}&\color{blue}{-26}&\color{orangered}{-57} \end{array} $$The solution is:
$$ \frac{ -5x^{4}+3x^{3}+x^{2}-5 }{ x-2 } = \color{blue}{-5x^{3}-7x^{2}-13x-26} \color{red}{~-~} \frac{ \color{red}{ 57 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&-5&3&1&0&-5\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ -5 }&3&1&0&-5\\& & & & & \\ \hline &\color{orangered}{-5}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&-5&3&1&0&-5\\& & \color{blue}{-10} & & & \\ \hline &\color{blue}{-5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}2&-5&\color{orangered}{ 3 }&1&0&-5\\& & \color{orangered}{-10} & & & \\ \hline &-5&\color{orangered}{-7}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -14 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&-5&3&1&0&-5\\& & -10& \color{blue}{-14} & & \\ \hline &-5&\color{blue}{-7}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -14 \right) } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrrr}2&-5&3&\color{orangered}{ 1 }&0&-5\\& & -10& \color{orangered}{-14} & & \\ \hline &-5&-7&\color{orangered}{-13}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -13 \right) } = \color{blue}{ -26 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&-5&3&1&0&-5\\& & -10& -14& \color{blue}{-26} & \\ \hline &-5&-7&\color{blue}{-13}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -26 \right) } = \color{orangered}{ -26 } $
$$ \begin{array}{c|rrrrr}2&-5&3&1&\color{orangered}{ 0 }&-5\\& & -10& -14& \color{orangered}{-26} & \\ \hline &-5&-7&-13&\color{orangered}{-26}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -26 \right) } = \color{blue}{ -52 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&-5&3&1&0&-5\\& & -10& -14& -26& \color{blue}{-52} \\ \hline &-5&-7&-13&\color{blue}{-26}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -52 \right) } = \color{orangered}{ -57 } $
$$ \begin{array}{c|rrrrr}2&-5&3&1&0&\color{orangered}{ -5 }\\& & -10& -14& -26& \color{orangered}{-52} \\ \hline &\color{blue}{-5}&\color{blue}{-7}&\color{blue}{-13}&\color{blue}{-26}&\color{orangered}{-57} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -5x^{3}-7x^{2}-13x-26 } $ with a remainder of $ \color{red}{ -57 } $.