Divide using synthetic division $$ ( 5x^{4}-7x^{3}+10x^{2}-12x+16 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&5&-7&10&-12&16\\& & 10& 6& 32& \color{black}{40} \\ \hline &\color{blue}{5}&\color{blue}{3}&\color{blue}{16}&\color{blue}{20}&\color{orangered}{56} \end{array} $$The solution is:
$$ \frac{ 5x^{4}-7x^{3}+10x^{2}-12x+16 }{ x-2 } = \color{blue}{5x^{3}+3x^{2}+16x+20} ~+~ \frac{ \color{red}{ 56 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&-7&10&-12&16\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 5 }&-7&10&-12&16\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&-7&10&-12&16\\& & \color{blue}{10} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 10 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}2&5&\color{orangered}{ -7 }&10&-12&16\\& & \color{orangered}{10} & & & \\ \hline &5&\color{orangered}{3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 3 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&-7&10&-12&16\\& & 10& \color{blue}{6} & & \\ \hline &5&\color{blue}{3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ 6 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrrr}2&5&-7&\color{orangered}{ 10 }&-12&16\\& & 10& \color{orangered}{6} & & \\ \hline &5&3&\color{orangered}{16}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 16 } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&-7&10&-12&16\\& & 10& 6& \color{blue}{32} & \\ \hline &5&3&\color{blue}{16}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 32 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrrr}2&5&-7&10&\color{orangered}{ -12 }&16\\& & 10& 6& \color{orangered}{32} & \\ \hline &5&3&16&\color{orangered}{20}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 20 } = \color{blue}{ 40 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&-7&10&-12&16\\& & 10& 6& 32& \color{blue}{40} \\ \hline &5&3&16&\color{blue}{20}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ 40 } = \color{orangered}{ 56 } $
$$ \begin{array}{c|rrrrr}2&5&-7&10&-12&\color{orangered}{ 16 }\\& & 10& 6& 32& \color{orangered}{40} \\ \hline &\color{blue}{5}&\color{blue}{3}&\color{blue}{16}&\color{blue}{20}&\color{orangered}{56} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{3}+3x^{2}+16x+20 } $ with a remainder of $ \color{red}{ 56 } $.