Divide using synthetic division $$ ( 2x^{3}+x^{2}-6 ) \div ( 3x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&2&1&0&-6\\& & 8& 36& \color{black}{144} \\ \hline &\color{blue}{2}&\color{blue}{9}&\color{blue}{36}&\color{orangered}{138} \end{array} $$The solution is:
$$ \frac{ 2x^{3}+x^{2}-6 }{ x-4 } = \color{blue}{2x^{2}+9x+36} ~+~ \frac{ \color{red}{ 138 } }{ x-4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{4}&2&1&0&-6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 2 }&1&0&-6\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 2 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&2&1&0&-6\\& & \color{blue}{8} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 8 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrr}4&2&\color{orangered}{ 1 }&0&-6\\& & \color{orangered}{8} & & \\ \hline &2&\color{orangered}{9}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 9 } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&2&1&0&-6\\& & 8& \color{blue}{36} & \\ \hline &2&\color{blue}{9}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 36 } = \color{orangered}{ 36 } $
$$ \begin{array}{c|rrrr}4&2&1&\color{orangered}{ 0 }&-6\\& & 8& \color{orangered}{36} & \\ \hline &2&9&\color{orangered}{36}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 36 } = \color{blue}{ 144 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&2&1&0&-6\\& & 8& 36& \color{blue}{144} \\ \hline &2&9&\color{blue}{36}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 144 } = \color{orangered}{ 138 } $
$$ \begin{array}{c|rrrr}4&2&1&0&\color{orangered}{ -6 }\\& & 8& 36& \color{orangered}{144} \\ \hline &\color{blue}{2}&\color{blue}{9}&\color{blue}{36}&\color{orangered}{138} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}+9x+36 } $ with a remainder of $ \color{red}{ 138 } $.