Check if $ x-3 $ is a factor of $ -8x^{3}+34 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&-8&0&0&34\\& & -24& -72& \color{black}{-216} \\ \hline &\color{blue}{-8}&\color{blue}{-24}&\color{blue}{-72}&\color{orangered}{-182} \end{array} $$Because the remainder $ \left( \color{red}{ -182 } \right) $ is not zero, we conclude that the $ x-3 $ is not a factor of $ -8x^{3}+34$.
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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-8&0&0&34\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ -8 }&0&0&34\\& & & & \\ \hline &\color{orangered}{-8}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-8&0&0&34\\& & \color{blue}{-24} & & \\ \hline &\color{blue}{-8}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -24 } $
$$ \begin{array}{c|rrrr}3&-8&\color{orangered}{ 0 }&0&34\\& & \color{orangered}{-24} & & \\ \hline &-8&\color{orangered}{-24}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -24 \right) } = \color{blue}{ -72 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-8&0&0&34\\& & -24& \color{blue}{-72} & \\ \hline &-8&\color{blue}{-24}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -72 \right) } = \color{orangered}{ -72 } $
$$ \begin{array}{c|rrrr}3&-8&0&\color{orangered}{ 0 }&34\\& & -24& \color{orangered}{-72} & \\ \hline &-8&-24&\color{orangered}{-72}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -72 \right) } = \color{blue}{ -216 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-8&0&0&34\\& & -24& -72& \color{blue}{-216} \\ \hline &-8&-24&\color{blue}{-72}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 34 } + \color{orangered}{ \left( -216 \right) } = \color{orangered}{ -182 } $
$$ \begin{array}{c|rrrr}3&-8&0&0&\color{orangered}{ 34 }\\& & -24& -72& \color{orangered}{-216} \\ \hline &\color{blue}{-8}&\color{blue}{-24}&\color{blue}{-72}&\color{orangered}{-182} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -182 }\right)$.