Divide using synthetic division $$ ( 3x^{3}+25x^{2}-28 ) \div ( x+10 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-10&3&25&0&-28\\& & -30& 50& \color{black}{-500} \\ \hline &\color{blue}{3}&\color{blue}{-5}&\color{blue}{50}&\color{orangered}{-528} \end{array} $$The solution is:
$$ \frac{ 3x^{3}+25x^{2}-28 }{ x+10 } = \color{blue}{3x^{2}-5x+50} \color{red}{~-~} \frac{ \color{red}{ 528 } }{ x+10 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 10 = 0 $ ( $ x = \color{blue}{ -10 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&3&25&0&-28\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-10&\color{orangered}{ 3 }&25&0&-28\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ 3 } = \color{blue}{ -30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&3&25&0&-28\\& & \color{blue}{-30} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 25 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}-10&3&\color{orangered}{ 25 }&0&-28\\& & \color{orangered}{-30} & & \\ \hline &3&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 50 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&3&25&0&-28\\& & -30& \color{blue}{50} & \\ \hline &3&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 50 } = \color{orangered}{ 50 } $
$$ \begin{array}{c|rrrr}-10&3&25&\color{orangered}{ 0 }&-28\\& & -30& \color{orangered}{50} & \\ \hline &3&-5&\color{orangered}{50}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ 50 } = \color{blue}{ -500 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&3&25&0&-28\\& & -30& 50& \color{blue}{-500} \\ \hline &3&-5&\color{blue}{50}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -28 } + \color{orangered}{ \left( -500 \right) } = \color{orangered}{ -528 } $
$$ \begin{array}{c|rrrr}-10&3&25&0&\color{orangered}{ -28 }\\& & -30& 50& \color{orangered}{-500} \\ \hline &\color{blue}{3}&\color{blue}{-5}&\color{blue}{50}&\color{orangered}{-528} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}-5x+50 } $ with a remainder of $ \color{red}{ -528 } $.