Check if $ x-3 $ is a factor of $ 2x^{3}-5x^{2}-8x+15 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&2&-5&-8&15\\& & 6& 3& \color{black}{-15} \\ \hline &\color{blue}{2}&\color{blue}{1}&\color{blue}{-5}&\color{orangered}{0} \end{array} $$Because the remainder equals zero, we conclude that the $ x-3 $ is a factor of the $ 2x^{3}-5x^{2}-8x+15 $.
Explanation
First we need to create a synthetic division table.
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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&2&-5&-8&15\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 2 }&-5&-8&15\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&2&-5&-8&15\\& & \color{blue}{6} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 6 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}3&2&\color{orangered}{ -5 }&-8&15\\& & \color{orangered}{6} & & \\ \hline &2&\color{orangered}{1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&2&-5&-8&15\\& & 6& \color{blue}{3} & \\ \hline &2&\color{blue}{1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 3 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}3&2&-5&\color{orangered}{ -8 }&15\\& & 6& \color{orangered}{3} & \\ \hline &2&1&\color{orangered}{-5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&2&-5&-8&15\\& & 6& 3& \color{blue}{-15} \\ \hline &2&1&\color{blue}{-5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}3&2&-5&-8&\color{orangered}{ 15 }\\& & 6& 3& \color{orangered}{-15} \\ \hline &\color{blue}{2}&\color{blue}{1}&\color{blue}{-5}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right)$.