Check if $ x+3 $ is a factor of $ 5x^{3}-3x^{2}-10x-124 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&5&-3&-10&-124\\& & -15& 54& \color{black}{-132} \\ \hline &\color{blue}{5}&\color{blue}{-18}&\color{blue}{44}&\color{orangered}{-256} \end{array} $$Because the remainder $ \left( \color{red}{ -256 } \right) $ is not zero, we conclude that the $ x+3 $ is not a factor of $ 5x^{3}-3x^{2}-10x-124$.
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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&5&-3&-10&-124\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 5 }&-3&-10&-124\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 5 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&5&-3&-10&-124\\& & \color{blue}{-15} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -18 } $
$$ \begin{array}{c|rrrr}-3&5&\color{orangered}{ -3 }&-10&-124\\& & \color{orangered}{-15} & & \\ \hline &5&\color{orangered}{-18}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -18 \right) } = \color{blue}{ 54 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&5&-3&-10&-124\\& & -15& \color{blue}{54} & \\ \hline &5&\color{blue}{-18}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 54 } = \color{orangered}{ 44 } $
$$ \begin{array}{c|rrrr}-3&5&-3&\color{orangered}{ -10 }&-124\\& & -15& \color{orangered}{54} & \\ \hline &5&-18&\color{orangered}{44}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 44 } = \color{blue}{ -132 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&5&-3&-10&-124\\& & -15& 54& \color{blue}{-132} \\ \hline &5&-18&\color{blue}{44}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -124 } + \color{orangered}{ \left( -132 \right) } = \color{orangered}{ -256 } $
$$ \begin{array}{c|rrrr}-3&5&-3&-10&\color{orangered}{ -124 }\\& & -15& 54& \color{orangered}{-132} \\ \hline &\color{blue}{5}&\color{blue}{-18}&\color{blue}{44}&\color{orangered}{-256} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -256 }\right)$.