Divide using synthetic division $$ ( 2x^{3}-13x^{2}-38x-24 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&2&-13&-38&-24\\& & 8& -20& \color{black}{-232} \\ \hline &\color{blue}{2}&\color{blue}{-5}&\color{blue}{-58}&\color{orangered}{-256} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-13x^{2}-38x-24 }{ x-4 } = \color{blue}{2x^{2}-5x-58} \color{red}{~-~} \frac{ \color{red}{ 256 } }{ x-4 } $$Explanation
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.
$$ \begin{array}{c|rrrr}\color{blue}{4}&2&-13&-38&-24\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 2 }&-13&-38&-24\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 2 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&2&-13&-38&-24\\& & \color{blue}{8} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 8 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}4&2&\color{orangered}{ -13 }&-38&-24\\& & \color{orangered}{8} & & \\ \hline &2&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&2&-13&-38&-24\\& & 8& \color{blue}{-20} & \\ \hline &2&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -38 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -58 } $
$$ \begin{array}{c|rrrr}4&2&-13&\color{orangered}{ -38 }&-24\\& & 8& \color{orangered}{-20} & \\ \hline &2&-5&\color{orangered}{-58}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -58 \right) } = \color{blue}{ -232 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&2&-13&-38&-24\\& & 8& -20& \color{blue}{-232} \\ \hline &2&-5&\color{blue}{-58}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -24 } + \color{orangered}{ \left( -232 \right) } = \color{orangered}{ -256 } $
$$ \begin{array}{c|rrrr}4&2&-13&-38&\color{orangered}{ -24 }\\& & 8& -20& \color{orangered}{-232} \\ \hline &\color{blue}{2}&\color{blue}{-5}&\color{blue}{-58}&\color{orangered}{-256} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-5x-58 } $ with a remainder of $ \color{red}{ -256 } $.