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