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