Divide using synthetic division $$ ( 3x^{3}+14x^{2}-10x-28 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&3&14&-10&-28\\& & -15& 5& \color{black}{25} \\ \hline &\color{blue}{3}&\color{blue}{-1}&\color{blue}{-5}&\color{orangered}{-3} \end{array} $$The solution is:
$$ \frac{ 3x^{3}+14x^{2}-10x-28 }{ x+5 } = \color{blue}{3x^{2}-x-5} \color{red}{~-~} \frac{ \color{red}{ 3 } }{ x+5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 5 = 0 $ ( $ x = \color{blue}{ -5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&3&14&-10&-28\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 3 }&14&-10&-28\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 3 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&3&14&-10&-28\\& & \color{blue}{-15} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}-5&3&\color{orangered}{ 14 }&-10&-28\\& & \color{orangered}{-15} & & \\ \hline &3&\color{orangered}{-1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&3&14&-10&-28\\& & -15& \color{blue}{5} & \\ \hline &3&\color{blue}{-1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 5 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}-5&3&14&\color{orangered}{ -10 }&-28\\& & -15& \color{orangered}{5} & \\ \hline &3&-1&\color{orangered}{-5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 25 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&3&14&-10&-28\\& & -15& 5& \color{blue}{25} \\ \hline &3&-1&\color{blue}{-5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -28 } + \color{orangered}{ 25 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}-5&3&14&-10&\color{orangered}{ -28 }\\& & -15& 5& \color{orangered}{25} \\ \hline &\color{blue}{3}&\color{blue}{-1}&\color{blue}{-5}&\color{orangered}{-3} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}-x-5 } $ with a remainder of $ \color{red}{ -3 } $.