Divide using synthetic division $$ ( 4x^{4}+12x^{3}-13x^{2}+9x-22 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&4&12&-13&9&-22\\& & -16& 16& -12& \color{black}{12} \\ \hline &\color{blue}{4}&\color{blue}{-4}&\color{blue}{3}&\color{blue}{-3}&\color{orangered}{-10} \end{array} $$The solution is:
$$ \frac{ 4x^{4}+12x^{3}-13x^{2}+9x-22 }{ x+4 } = \color{blue}{4x^{3}-4x^{2}+3x-3} \color{red}{~-~} \frac{ \color{red}{ 10 } }{ x+4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 4 = 0 $ ( $ x = \color{blue}{ -4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&4&12&-13&9&-22\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 4 }&12&-13&9&-22\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 4 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&4&12&-13&9&-22\\& & \color{blue}{-16} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}-4&4&\color{orangered}{ 12 }&-13&9&-22\\& & \color{orangered}{-16} & & & \\ \hline &4&\color{orangered}{-4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&4&12&-13&9&-22\\& & -16& \color{blue}{16} & & \\ \hline &4&\color{blue}{-4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 16 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}-4&4&12&\color{orangered}{ -13 }&9&-22\\& & -16& \color{orangered}{16} & & \\ \hline &4&-4&\color{orangered}{3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 3 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&4&12&-13&9&-22\\& & -16& 16& \color{blue}{-12} & \\ \hline &4&-4&\color{blue}{3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-4&4&12&-13&\color{orangered}{ 9 }&-22\\& & -16& 16& \color{orangered}{-12} & \\ \hline &4&-4&3&\color{orangered}{-3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&4&12&-13&9&-22\\& & -16& 16& -12& \color{blue}{12} \\ \hline &4&-4&3&\color{blue}{-3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -22 } + \color{orangered}{ 12 } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}-4&4&12&-13&9&\color{orangered}{ -22 }\\& & -16& 16& -12& \color{orangered}{12} \\ \hline &\color{blue}{4}&\color{blue}{-4}&\color{blue}{3}&\color{blue}{-3}&\color{orangered}{-10} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}-4x^{2}+3x-3 } $ with a remainder of $ \color{red}{ -10 } $.