Divide using synthetic division $$ ( 2x^{3}-5x^{2}-28x+15 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&2&-5&-28&15\\& & 4& -2& \color{black}{-60} \\ \hline &\color{blue}{2}&\color{blue}{-1}&\color{blue}{-30}&\color{orangered}{-45} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-5x^{2}-28x+15 }{ x-2 } = \color{blue}{2x^{2}-x-30} \color{red}{~-~} \frac{ \color{red}{ 45 } }{ x-2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{2}&2&-5&-28&15\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 2 }&-5&-28&15\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&2&-5&-28&15\\& & \color{blue}{4} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 4 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}2&2&\color{orangered}{ -5 }&-28&15\\& & \color{orangered}{4} & & \\ \hline &2&\color{orangered}{-1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&2&-5&-28&15\\& & 4& \color{blue}{-2} & \\ \hline &2&\color{blue}{-1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -28 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -30 } $
$$ \begin{array}{c|rrrr}2&2&-5&\color{orangered}{ -28 }&15\\& & 4& \color{orangered}{-2} & \\ \hline &2&-1&\color{orangered}{-30}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -30 \right) } = \color{blue}{ -60 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&2&-5&-28&15\\& & 4& -2& \color{blue}{-60} \\ \hline &2&-1&\color{blue}{-30}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -60 \right) } = \color{orangered}{ -45 } $
$$ \begin{array}{c|rrrr}2&2&-5&-28&\color{orangered}{ 15 }\\& & 4& -2& \color{orangered}{-60} \\ \hline &\color{blue}{2}&\color{blue}{-1}&\color{blue}{-30}&\color{orangered}{-45} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-x-30 } $ with a remainder of $ \color{red}{ -45 } $.