Divide using synthetic division $$ ( 5x^{4}-14x^{3}+77x^{2}-224x-48 ) \div ( 3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}0&5&-14&77&-224&-48\\& & 0& 0& 0& \color{black}{0} \\ \hline &\color{blue}{5}&\color{blue}{-14}&\color{blue}{77}&\color{blue}{-224}&\color{orangered}{-48} \end{array} $$The solution is:
$$ \frac{ 5x^{4}-14x^{3}+77x^{2}-224x-48 }{ x } = \color{blue}{5x^{3}-14x^{2}+77x-224} \color{red}{~-~} \frac{ \color{red}{ 48 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&5&-14&77&-224&-48\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}0&\color{orangered}{ 5 }&-14&77&-224&-48\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 5 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&5&-14&77&-224&-48\\& & \color{blue}{0} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 0 } = \color{orangered}{ -14 } $
$$ \begin{array}{c|rrrrr}0&5&\color{orangered}{ -14 }&77&-224&-48\\& & \color{orangered}{0} & & & \\ \hline &5&\color{orangered}{-14}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -14 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&5&-14&77&-224&-48\\& & 0& \color{blue}{0} & & \\ \hline &5&\color{blue}{-14}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 77 } + \color{orangered}{ 0 } = \color{orangered}{ 77 } $
$$ \begin{array}{c|rrrrr}0&5&-14&\color{orangered}{ 77 }&-224&-48\\& & 0& \color{orangered}{0} & & \\ \hline &5&-14&\color{orangered}{77}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 77 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&5&-14&77&-224&-48\\& & 0& 0& \color{blue}{0} & \\ \hline &5&-14&\color{blue}{77}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -224 } + \color{orangered}{ 0 } = \color{orangered}{ -224 } $
$$ \begin{array}{c|rrrrr}0&5&-14&77&\color{orangered}{ -224 }&-48\\& & 0& 0& \color{orangered}{0} & \\ \hline &5&-14&77&\color{orangered}{-224}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -224 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&5&-14&77&-224&-48\\& & 0& 0& 0& \color{blue}{0} \\ \hline &5&-14&77&\color{blue}{-224}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -48 } + \color{orangered}{ 0 } = \color{orangered}{ -48 } $
$$ \begin{array}{c|rrrrr}0&5&-14&77&-224&\color{orangered}{ -48 }\\& & 0& 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{5}&\color{blue}{-14}&\color{blue}{77}&\color{blue}{-224}&\color{orangered}{-48} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{3}-14x^{2}+77x-224 } $ with a remainder of $ \color{red}{ -48 } $.