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