Divide using synthetic division $$ ( 4x^{3}+7x^{2}-10x+3 ) \div ( 2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}0&4&7&-10&3\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{4}&\color{blue}{7}&\color{blue}{-10}&\color{orangered}{3} \end{array} $$The solution is:
$$ \frac{ 4x^{3}+7x^{2}-10x+3 }{ x } = \color{blue}{4x^{2}+7x-10} ~+~ \frac{ \color{red}{ 3 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrr}\color{blue}{0}&4&7&-10&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}0&\color{orangered}{ 4 }&7&-10&3\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&4&7&-10&3\\& & \color{blue}{0} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 0 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}0&4&\color{orangered}{ 7 }&-10&3\\& & \color{orangered}{0} & & \\ \hline &4&\color{orangered}{7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 7 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&4&7&-10&3\\& & 0& \color{blue}{0} & \\ \hline &4&\color{blue}{7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 0 } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrr}0&4&7&\color{orangered}{ -10 }&3\\& & 0& \color{orangered}{0} & \\ \hline &4&7&\color{orangered}{-10}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&4&7&-10&3\\& & 0& 0& \color{blue}{0} \\ \hline &4&7&\color{blue}{-10}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 0 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}0&4&7&-10&\color{orangered}{ 3 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{4}&\color{blue}{7}&\color{blue}{-10}&\color{orangered}{3} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+7x-10 } $ with a remainder of $ \color{red}{ 3 } $.