Divide using synthetic division $$ ( 2x^{3}-14x^{2}+15x-18 ) \div ( x-6 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}6&2&-14&15&-18\\& & 12& -12& \color{black}{18} \\ \hline &\color{blue}{2}&\color{blue}{-2}&\color{blue}{3}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-14x^{2}+15x-18 }{ x-6 } = \color{blue}{2x^{2}-2x+3} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -6 = 0 $ ( $ x = \color{blue}{ 6 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{6}&2&-14&15&-18\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}6&\color{orangered}{ 2 }&-14&15&-18\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 2 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{6}&2&-14&15&-18\\& & \color{blue}{12} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 12 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}6&2&\color{orangered}{ -14 }&15&-18\\& & \color{orangered}{12} & & \\ \hline &2&\color{orangered}{-2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{6}&2&-14&15&-18\\& & 12& \color{blue}{-12} & \\ \hline &2&\color{blue}{-2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}6&2&-14&\color{orangered}{ 15 }&-18\\& & 12& \color{orangered}{-12} & \\ \hline &2&-2&\color{orangered}{3}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 3 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{6}&2&-14&15&-18\\& & 12& -12& \color{blue}{18} \\ \hline &2&-2&\color{blue}{3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 18 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}6&2&-14&15&\color{orangered}{ -18 }\\& & 12& -12& \color{orangered}{18} \\ \hline &\color{blue}{2}&\color{blue}{-2}&\color{blue}{3}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-2x+3 } $ with a remainder of $ \color{red}{ 0 } $.