Check if $ x+3 $ is a factor of $ 8x^{3}-10x^{2}-x+3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&8&-10&-1&3\\& & -24& 102& \color{black}{-303} \\ \hline &\color{blue}{8}&\color{blue}{-34}&\color{blue}{101}&\color{orangered}{-300} \end{array} $$Because the remainder $ \left( \color{red}{ -300 } \right) $ is not zero, we conclude that the $ x+3 $ is not a factor of $ 8x^{3}-10x^{2}-x+3$.
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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&8&-10&-1&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 8 }&-10&-1&3\\& & & & \\ \hline &\color{orangered}{8}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 8 } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&8&-10&-1&3\\& & \color{blue}{-24} & & \\ \hline &\color{blue}{8}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -34 } $
$$ \begin{array}{c|rrrr}-3&8&\color{orangered}{ -10 }&-1&3\\& & \color{orangered}{-24} & & \\ \hline &8&\color{orangered}{-34}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -34 \right) } = \color{blue}{ 102 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&8&-10&-1&3\\& & -24& \color{blue}{102} & \\ \hline &8&\color{blue}{-34}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 102 } = \color{orangered}{ 101 } $
$$ \begin{array}{c|rrrr}-3&8&-10&\color{orangered}{ -1 }&3\\& & -24& \color{orangered}{102} & \\ \hline &8&-34&\color{orangered}{101}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 101 } = \color{blue}{ -303 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&8&-10&-1&3\\& & -24& 102& \color{blue}{-303} \\ \hline &8&-34&\color{blue}{101}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -303 \right) } = \color{orangered}{ -300 } $
$$ \begin{array}{c|rrrr}-3&8&-10&-1&\color{orangered}{ 3 }\\& & -24& 102& \color{orangered}{-303} \\ \hline &\color{blue}{8}&\color{blue}{-34}&\color{blue}{101}&\color{orangered}{-300} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -300 }\right)$.