Divide using synthetic division $$ ( 2x^{4}-11x^{3}+15x^{2}+6x-18 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&2&-11&15&6&-18\\& & 6& -15& 0& \color{black}{18} \\ \hline &\color{blue}{2}&\color{blue}{-5}&\color{blue}{0}&\color{blue}{6}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-11x^{3}+15x^{2}+6x-18 }{ x-3 } = \color{blue}{2x^{3}-5x^{2}+6} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-11&15&6&-18\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 2 }&-11&15&6&-18\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-11&15&6&-18\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 6 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}3&2&\color{orangered}{ -11 }&15&6&-18\\& & \color{orangered}{6} & & & \\ \hline &2&\color{orangered}{-5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-11&15&6&-18\\& & 6& \color{blue}{-15} & & \\ \hline &2&\color{blue}{-5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}3&2&-11&\color{orangered}{ 15 }&6&-18\\& & 6& \color{orangered}{-15} & & \\ \hline &2&-5&\color{orangered}{0}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-11&15&6&-18\\& & 6& -15& \color{blue}{0} & \\ \hline &2&-5&\color{blue}{0}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 0 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}3&2&-11&15&\color{orangered}{ 6 }&-18\\& & 6& -15& \color{orangered}{0} & \\ \hline &2&-5&0&\color{orangered}{6}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 6 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-11&15&6&-18\\& & 6& -15& 0& \color{blue}{18} \\ \hline &2&-5&0&\color{blue}{6}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 18 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}3&2&-11&15&6&\color{orangered}{ -18 }\\& & 6& -15& 0& \color{orangered}{18} \\ \hline &\color{blue}{2}&\color{blue}{-5}&\color{blue}{0}&\color{blue}{6}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-5x^{2}+6 } $ with a remainder of $ \color{red}{ 0 } $.