Divide using synthetic division $$ ( x^{4}+2x^{3}-13x^{2}-14x+24 ) \div ( 2x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&1&2&-13&-14&24\\& & 1& 3& -10& \color{black}{-24} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{-10}&\color{blue}{-24}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{4}+2x^{3}-13x^{2}-14x+24 }{ x-1 } = \color{blue}{x^{3}+3x^{2}-10x-24} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&2&-13&-14&24\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 1 }&2&-13&-14&24\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 1 } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&2&-13&-14&24\\& & \color{blue}{1} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 1 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}1&1&\color{orangered}{ 2 }&-13&-14&24\\& & \color{orangered}{1} & & & \\ \hline &1&\color{orangered}{3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 3 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&2&-13&-14&24\\& & 1& \color{blue}{3} & & \\ \hline &1&\color{blue}{3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 3 } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}1&1&2&\color{orangered}{ -13 }&-14&24\\& & 1& \color{orangered}{3} & & \\ \hline &1&3&\color{orangered}{-10}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&2&-13&-14&24\\& & 1& 3& \color{blue}{-10} & \\ \hline &1&3&\color{blue}{-10}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -24 } $
$$ \begin{array}{c|rrrrr}1&1&2&-13&\color{orangered}{ -14 }&24\\& & 1& 3& \color{orangered}{-10} & \\ \hline &1&3&-10&\color{orangered}{-24}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -24 \right) } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&2&-13&-14&24\\& & 1& 3& -10& \color{blue}{-24} \\ \hline &1&3&-10&\color{blue}{-24}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 24 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}1&1&2&-13&-14&\color{orangered}{ 24 }\\& & 1& 3& -10& \color{orangered}{-24} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{-10}&\color{blue}{-24}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+3x^{2}-10x-24 } $ with a remainder of $ \color{red}{ 0 } $.