Divide using synthetic division $$ ( 4x^{3}+32x^{2}-77x+30 ) \div ( x+10 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-10&4&32&-77&30\\& & -40& 80& \color{black}{-30} \\ \hline &\color{blue}{4}&\color{blue}{-8}&\color{blue}{3}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 4x^{3}+32x^{2}-77x+30 }{ x+10 } = \color{blue}{4x^{2}-8x+3} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&4&32&-77&30\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-10&\color{orangered}{ 4 }&32&-77&30\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ 4 } = \color{blue}{ -40 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&4&32&-77&30\\& & \color{blue}{-40} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 32 } + \color{orangered}{ \left( -40 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}-10&4&\color{orangered}{ 32 }&-77&30\\& & \color{orangered}{-40} & & \\ \hline &4&\color{orangered}{-8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 80 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&4&32&-77&30\\& & -40& \color{blue}{80} & \\ \hline &4&\color{blue}{-8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -77 } + \color{orangered}{ 80 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}-10&4&32&\color{orangered}{ -77 }&30\\& & -40& \color{orangered}{80} & \\ \hline &4&-8&\color{orangered}{3}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ 3 } = \color{blue}{ -30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&4&32&-77&30\\& & -40& 80& \color{blue}{-30} \\ \hline &4&-8&\color{blue}{3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 30 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-10&4&32&-77&\color{orangered}{ 30 }\\& & -40& 80& \color{orangered}{-30} \\ \hline &\color{blue}{4}&\color{blue}{-8}&\color{blue}{3}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-8x+3 } $ with a remainder of $ \color{red}{ 0 } $.