Divide using synthetic division $$ ( -4x^{3}+23x^{2}+32x+22 ) \div ( x-7 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}7&-4&23&32&22\\& & -28& -35& \color{black}{-21} \\ \hline &\color{blue}{-4}&\color{blue}{-5}&\color{blue}{-3}&\color{orangered}{1} \end{array} $$The solution is:
$$ \frac{ -4x^{3}+23x^{2}+32x+22 }{ x-7 } = \color{blue}{-4x^{2}-5x-3} ~+~ \frac{ \color{red}{ 1 } }{ x-7 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{7}&-4&23&32&22\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}7&\color{orangered}{ -4 }&23&32&22\\& & & & \\ \hline &\color{orangered}{-4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -28 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&-4&23&32&22\\& & \color{blue}{-28} & & \\ \hline &\color{blue}{-4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 23 } + \color{orangered}{ \left( -28 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}7&-4&\color{orangered}{ 23 }&32&22\\& & \color{orangered}{-28} & & \\ \hline &-4&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -35 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&-4&23&32&22\\& & -28& \color{blue}{-35} & \\ \hline &-4&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 32 } + \color{orangered}{ \left( -35 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}7&-4&23&\color{orangered}{ 32 }&22\\& & -28& \color{orangered}{-35} & \\ \hline &-4&-5&\color{orangered}{-3}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -21 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&-4&23&32&22\\& & -28& -35& \color{blue}{-21} \\ \hline &-4&-5&\color{blue}{-3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 22 } + \color{orangered}{ \left( -21 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}7&-4&23&32&\color{orangered}{ 22 }\\& & -28& -35& \color{orangered}{-21} \\ \hline &\color{blue}{-4}&\color{blue}{-5}&\color{blue}{-3}&\color{orangered}{1} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -4x^{2}-5x-3 } $ with a remainder of $ \color{red}{ 1 } $.