Divide using synthetic division $$ ( 4x^{4}-15x^{3}-24x^{2}+x-30 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&4&-15&-24&1&-30\\& & 20& 25& 5& \color{black}{30} \\ \hline &\color{blue}{4}&\color{blue}{5}&\color{blue}{1}&\color{blue}{6}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 4x^{4}-15x^{3}-24x^{2}+x-30 }{ x-5 } = \color{blue}{4x^{3}+5x^{2}+x+6} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&4&-15&-24&1&-30\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 4 }&-15&-24&1&-30\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 4 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&4&-15&-24&1&-30\\& & \color{blue}{20} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 20 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}5&4&\color{orangered}{ -15 }&-24&1&-30\\& & \color{orangered}{20} & & & \\ \hline &4&\color{orangered}{5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 5 } = \color{blue}{ 25 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&4&-15&-24&1&-30\\& & 20& \color{blue}{25} & & \\ \hline &4&\color{blue}{5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -24 } + \color{orangered}{ 25 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}5&4&-15&\color{orangered}{ -24 }&1&-30\\& & 20& \color{orangered}{25} & & \\ \hline &4&5&\color{orangered}{1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&4&-15&-24&1&-30\\& & 20& 25& \color{blue}{5} & \\ \hline &4&5&\color{blue}{1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 5 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}5&4&-15&-24&\color{orangered}{ 1 }&-30\\& & 20& 25& \color{orangered}{5} & \\ \hline &4&5&1&\color{orangered}{6}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 6 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&4&-15&-24&1&-30\\& & 20& 25& 5& \color{blue}{30} \\ \hline &4&5&1&\color{blue}{6}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -30 } + \color{orangered}{ 30 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}5&4&-15&-24&1&\color{orangered}{ -30 }\\& & 20& 25& 5& \color{orangered}{30} \\ \hline &\color{blue}{4}&\color{blue}{5}&\color{blue}{1}&\color{blue}{6}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}+5x^{2}+x+6 } $ with a remainder of $ \color{red}{ 0 } $.