Divide using synthetic division $$ ( x^{4}-23x^{2}-50 ) \div ( 5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}0&1&0&-23&0&-50\\& & 0& 0& 0& \color{black}{0} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{-23}&\color{blue}{0}&\color{orangered}{-50} \end{array} $$The solution is:
$$ \frac{ x^{4}-23x^{2}-50 }{ x } = \color{blue}{x^{3}-23x} \color{red}{~-~} \frac{ \color{red}{ 50 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&1&0&-23&0&-50\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}0&\color{orangered}{ 1 }&0&-23&0&-50\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 1 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&1&0&-23&0&-50\\& & \color{blue}{0} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}0&1&\color{orangered}{ 0 }&-23&0&-50\\& & \color{orangered}{0} & & & \\ \hline &1&\color{orangered}{0}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&1&0&-23&0&-50\\& & 0& \color{blue}{0} & & \\ \hline &1&\color{blue}{0}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -23 } + \color{orangered}{ 0 } = \color{orangered}{ -23 } $
$$ \begin{array}{c|rrrrr}0&1&0&\color{orangered}{ -23 }&0&-50\\& & 0& \color{orangered}{0} & & \\ \hline &1&0&\color{orangered}{-23}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -23 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&1&0&-23&0&-50\\& & 0& 0& \color{blue}{0} & \\ \hline &1&0&\color{blue}{-23}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}0&1&0&-23&\color{orangered}{ 0 }&-50\\& & 0& 0& \color{orangered}{0} & \\ \hline &1&0&-23&\color{orangered}{0}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&1&0&-23&0&-50\\& & 0& 0& 0& \color{blue}{0} \\ \hline &1&0&-23&\color{blue}{0}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -50 } + \color{orangered}{ 0 } = \color{orangered}{ -50 } $
$$ \begin{array}{c|rrrrr}0&1&0&-23&0&\color{orangered}{ -50 }\\& & 0& 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{-23}&\color{blue}{0}&\color{orangered}{-50} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-23x } $ with a remainder of $ \color{red}{ -50 } $.