Check if $ x+4 $ is a factor of $ x^{3}+8x^{2}+10x-24 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-4&1&8&10&-24\\& & -4& -16& \color{black}{24} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{-6}&\color{orangered}{0} \end{array} $$Because the remainder equals zero, we conclude that the $ x+4 $ is a factor of the $ x^{3}+8x^{2}+10x-24 $.
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 + 4 = 0 $ ( $ x = \color{blue}{ -4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&8&10&-24\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-4&\color{orangered}{ 1 }&8&10&-24\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 1 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&8&10&-24\\& & \color{blue}{-4} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}-4&1&\color{orangered}{ 8 }&10&-24\\& & \color{orangered}{-4} & & \\ \hline &1&\color{orangered}{4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 4 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&8&10&-24\\& & -4& \color{blue}{-16} & \\ \hline &1&\color{blue}{4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}-4&1&8&\color{orangered}{ 10 }&-24\\& & -4& \color{orangered}{-16} & \\ \hline &1&4&\color{orangered}{-6}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&8&10&-24\\& & -4& -16& \color{blue}{24} \\ \hline &1&4&\color{blue}{-6}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -24 } + \color{orangered}{ 24 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-4&1&8&10&\color{orangered}{ -24 }\\& & -4& -16& \color{orangered}{24} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{-6}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right)$.