Divide using synthetic division $$ ( 5x^{4}+14x^{3}+13x+12 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&5&14&0&13&12\\& & -15& 3& -9& \color{black}{-12} \\ \hline &\color{blue}{5}&\color{blue}{-1}&\color{blue}{3}&\color{blue}{4}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 5x^{4}+14x^{3}+13x+12 }{ x+3 } = \color{blue}{5x^{3}-x^{2}+3x+4} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&5&14&0&13&12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 5 }&14&0&13&12\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 5 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&5&14&0&13&12\\& & \color{blue}{-15} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}-3&5&\color{orangered}{ 14 }&0&13&12\\& & \color{orangered}{-15} & & & \\ \hline &5&\color{orangered}{-1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&5&14&0&13&12\\& & -15& \color{blue}{3} & & \\ \hline &5&\color{blue}{-1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 3 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}-3&5&14&\color{orangered}{ 0 }&13&12\\& & -15& \color{orangered}{3} & & \\ \hline &5&-1&\color{orangered}{3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 3 } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&5&14&0&13&12\\& & -15& 3& \color{blue}{-9} & \\ \hline &5&-1&\color{blue}{3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}-3&5&14&0&\color{orangered}{ 13 }&12\\& & -15& 3& \color{orangered}{-9} & \\ \hline &5&-1&3&\color{orangered}{4}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 4 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&5&14&0&13&12\\& & -15& 3& -9& \color{blue}{-12} \\ \hline &5&-1&3&\color{blue}{4}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-3&5&14&0&13&\color{orangered}{ 12 }\\& & -15& 3& -9& \color{orangered}{-12} \\ \hline &\color{blue}{5}&\color{blue}{-1}&\color{blue}{3}&\color{blue}{4}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{3}-x^{2}+3x+4 } $ with a remainder of $ \color{red}{ 0 } $.