Check if $ x-3 $ is a factor of $ -4x^{3}+4x^{2}+15x $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&-4&4&15&0\\& & -12& -24& \color{black}{-27} \\ \hline &\color{blue}{-4}&\color{blue}{-8}&\color{blue}{-9}&\color{orangered}{-27} \end{array} $$Because the remainder $ \left( \color{red}{ -27 } \right) $ is not zero, we conclude that the $ x-3 $ is not a factor of $ -4x^{3}+4x^{2}+15x$.
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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-4&4&15&0\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ -4 }&4&15&0\\& & & & \\ \hline &\color{orangered}{-4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-4&4&15&0\\& & \color{blue}{-12} & & \\ \hline &\color{blue}{-4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}3&-4&\color{orangered}{ 4 }&15&0\\& & \color{orangered}{-12} & & \\ \hline &-4&\color{orangered}{-8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-4&4&15&0\\& & -12& \color{blue}{-24} & \\ \hline &-4&\color{blue}{-8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrr}3&-4&4&\color{orangered}{ 15 }&0\\& & -12& \color{orangered}{-24} & \\ \hline &-4&-8&\color{orangered}{-9}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ -27 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-4&4&15&0\\& & -12& -24& \color{blue}{-27} \\ \hline &-4&-8&\color{blue}{-9}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -27 \right) } = \color{orangered}{ -27 } $
$$ \begin{array}{c|rrrr}3&-4&4&15&\color{orangered}{ 0 }\\& & -12& -24& \color{orangered}{-27} \\ \hline &\color{blue}{-4}&\color{blue}{-8}&\color{blue}{-9}&\color{orangered}{-27} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -27 }\right)$.