Divide using synthetic division $$ ( 4x^{3}+10x^{2}+13x+9 ) \div ( x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&4&10&13&9\\& & -4& -6& \color{black}{-7} \\ \hline &\color{blue}{4}&\color{blue}{6}&\color{blue}{7}&\color{orangered}{2} \end{array} $$The solution is:
$$ \frac{ 4x^{3}+10x^{2}+13x+9 }{ x+1 } = \color{blue}{4x^{2}+6x+7} ~+~ \frac{ \color{red}{ 2 } }{ x+1 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 1 = 0 $ ( $ x = \color{blue}{ -1 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&4&10&13&9\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 4 }&10&13&9\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 4 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&4&10&13&9\\& & \color{blue}{-4} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}-1&4&\color{orangered}{ 10 }&13&9\\& & \color{orangered}{-4} & & \\ \hline &4&\color{orangered}{6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 6 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&4&10&13&9\\& & -4& \color{blue}{-6} & \\ \hline &4&\color{blue}{6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}-1&4&10&\color{orangered}{ 13 }&9\\& & -4& \color{orangered}{-6} & \\ \hline &4&6&\color{orangered}{7}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 7 } = \color{blue}{ -7 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&4&10&13&9\\& & -4& -6& \color{blue}{-7} \\ \hline &4&6&\color{blue}{7}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -7 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}-1&4&10&13&\color{orangered}{ 9 }\\& & -4& -6& \color{orangered}{-7} \\ \hline &\color{blue}{4}&\color{blue}{6}&\color{blue}{7}&\color{orangered}{2} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+6x+7 } $ with a remainder of $ \color{red}{ 2 } $.