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