Divide using synthetic division $$ ( 2x^{3}+4x^{2}-28x+3 ) \div ( 2x-6 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}6&2&4&-28&3\\& & 12& 96& \color{black}{408} \\ \hline &\color{blue}{2}&\color{blue}{16}&\color{blue}{68}&\color{orangered}{411} \end{array} $$The solution is:
$$ \frac{ 2x^{3}+4x^{2}-28x+3 }{ x-6 } = \color{blue}{2x^{2}+16x+68} ~+~ \frac{ \color{red}{ 411 } }{ x-6 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{6}&2&4&-28&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}6&\color{orangered}{ 2 }&4&-28&3\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{6}&2&4&-28&3\\& & \color{blue}{12} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 12 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrr}6&2&\color{orangered}{ 4 }&-28&3\\& & \color{orangered}{12} & & \\ \hline &2&\color{orangered}{16}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 16 } = \color{blue}{ 96 } $.
$$ \begin{array}{c|rrrr}\color{blue}{6}&2&4&-28&3\\& & 12& \color{blue}{96} & \\ \hline &2&\color{blue}{16}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -28 } + \color{orangered}{ 96 } = \color{orangered}{ 68 } $
$$ \begin{array}{c|rrrr}6&2&4&\color{orangered}{ -28 }&3\\& & 12& \color{orangered}{96} & \\ \hline &2&16&\color{orangered}{68}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 68 } = \color{blue}{ 408 } $.
$$ \begin{array}{c|rrrr}\color{blue}{6}&2&4&-28&3\\& & 12& 96& \color{blue}{408} \\ \hline &2&16&\color{blue}{68}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 408 } = \color{orangered}{ 411 } $
$$ \begin{array}{c|rrrr}6&2&4&-28&\color{orangered}{ 3 }\\& & 12& 96& \color{orangered}{408} \\ \hline &\color{blue}{2}&\color{blue}{16}&\color{blue}{68}&\color{orangered}{411} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}+16x+68 } $ with a remainder of $ \color{red}{ 411 } $.