Divide using synthetic division $$ ( 9x^{3}-128x^{2}+478x-100 ) \div ( x+7 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-7&9&-128&478&-100\\& & -63& 1337& \color{black}{-12705} \\ \hline &\color{blue}{9}&\color{blue}{-191}&\color{blue}{1815}&\color{orangered}{-12805} \end{array} $$The solution is:
$$ \frac{ 9x^{3}-128x^{2}+478x-100 }{ x+7 } = \color{blue}{9x^{2}-191x+1815} \color{red}{~-~} \frac{ \color{red}{ 12805 } }{ x+7 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 7 = 0 $ ( $ x = \color{blue}{ -7 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-7}&9&-128&478&-100\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-7&\color{orangered}{ 9 }&-128&478&-100\\& & & & \\ \hline &\color{orangered}{9}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -7 } \cdot \color{blue}{ 9 } = \color{blue}{ -63 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-7}&9&-128&478&-100\\& & \color{blue}{-63} & & \\ \hline &\color{blue}{9}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -128 } + \color{orangered}{ \left( -63 \right) } = \color{orangered}{ -191 } $
$$ \begin{array}{c|rrrr}-7&9&\color{orangered}{ -128 }&478&-100\\& & \color{orangered}{-63} & & \\ \hline &9&\color{orangered}{-191}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -7 } \cdot \color{blue}{ \left( -191 \right) } = \color{blue}{ 1337 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-7}&9&-128&478&-100\\& & -63& \color{blue}{1337} & \\ \hline &9&\color{blue}{-191}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 478 } + \color{orangered}{ 1337 } = \color{orangered}{ 1815 } $
$$ \begin{array}{c|rrrr}-7&9&-128&\color{orangered}{ 478 }&-100\\& & -63& \color{orangered}{1337} & \\ \hline &9&-191&\color{orangered}{1815}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -7 } \cdot \color{blue}{ 1815 } = \color{blue}{ -12705 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-7}&9&-128&478&-100\\& & -63& 1337& \color{blue}{-12705} \\ \hline &9&-191&\color{blue}{1815}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -100 } + \color{orangered}{ \left( -12705 \right) } = \color{orangered}{ -12805 } $
$$ \begin{array}{c|rrrr}-7&9&-128&478&\color{orangered}{ -100 }\\& & -63& 1337& \color{orangered}{-12705} \\ \hline &\color{blue}{9}&\color{blue}{-191}&\color{blue}{1815}&\color{orangered}{-12805} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 9x^{2}-191x+1815 } $ with a remainder of $ \color{red}{ -12805 } $.