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