Divide using synthetic division $$ ( 9x^{3}+90x^{2}-2 ) \div ( x+10 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-10&9&90&0&-2\\& & -90& 0& \color{black}{0} \\ \hline &\color{blue}{9}&\color{blue}{0}&\color{blue}{0}&\color{orangered}{-2} \end{array} $$The solution is:
$$ \frac{ 9x^{3}+90x^{2}-2 }{ x+10 } = \color{blue}{9x^{2}} \color{red}{~-~} \frac{ \color{red}{ 2 } }{ x+10 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&9&90&0&-2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-10&\color{orangered}{ 9 }&90&0&-2\\& & & & \\ \hline &\color{orangered}{9}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ 9 } = \color{blue}{ -90 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&9&90&0&-2\\& & \color{blue}{-90} & & \\ \hline &\color{blue}{9}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 90 } + \color{orangered}{ \left( -90 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-10&9&\color{orangered}{ 90 }&0&-2\\& & \color{orangered}{-90} & & \\ \hline &9&\color{orangered}{0}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&9&90&0&-2\\& & -90& \color{blue}{0} & \\ \hline &9&\color{blue}{0}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-10&9&90&\color{orangered}{ 0 }&-2\\& & -90& \color{orangered}{0} & \\ \hline &9&0&\color{orangered}{0}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&9&90&0&-2\\& & -90& 0& \color{blue}{0} \\ \hline &9&0&\color{blue}{0}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 0 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}-10&9&90&0&\color{orangered}{ -2 }\\& & -90& 0& \color{orangered}{0} \\ \hline &\color{blue}{9}&\color{blue}{0}&\color{blue}{0}&\color{orangered}{-2} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 9x^{2} } $ with a remainder of $ \color{red}{ -2 } $.