Divide using synthetic division $$ ( 2x^{3}-3x^{2}+5x-14 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&2&-3&5&-14\\& & -4& 14& \color{black}{-38} \\ \hline &\color{blue}{2}&\color{blue}{-7}&\color{blue}{19}&\color{orangered}{-52} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-3x^{2}+5x-14 }{ x+2 } = \color{blue}{2x^{2}-7x+19} \color{red}{~-~} \frac{ \color{red}{ 52 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-3&5&-14\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 2 }&-3&5&-14\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 2 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-3&5&-14\\& & \color{blue}{-4} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}-2&2&\color{orangered}{ -3 }&5&-14\\& & \color{orangered}{-4} & & \\ \hline &2&\color{orangered}{-7}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-3&5&-14\\& & -4& \color{blue}{14} & \\ \hline &2&\color{blue}{-7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 14 } = \color{orangered}{ 19 } $
$$ \begin{array}{c|rrrr}-2&2&-3&\color{orangered}{ 5 }&-14\\& & -4& \color{orangered}{14} & \\ \hline &2&-7&\color{orangered}{19}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 19 } = \color{blue}{ -38 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-3&5&-14\\& & -4& 14& \color{blue}{-38} \\ \hline &2&-7&\color{blue}{19}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ \left( -38 \right) } = \color{orangered}{ -52 } $
$$ \begin{array}{c|rrrr}-2&2&-3&5&\color{orangered}{ -14 }\\& & -4& 14& \color{orangered}{-38} \\ \hline &\color{blue}{2}&\color{blue}{-7}&\color{blue}{19}&\color{orangered}{-52} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-7x+19 } $ with a remainder of $ \color{red}{ -52 } $.