Divide using synthetic division $$ ( x^{4}-25x^{2}+144 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&1&0&-25&0&144\\& & -3& 9& 48& \color{black}{-144} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{-16}&\color{blue}{48}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{4}-25x^{2}+144 }{ x+3 } = \color{blue}{x^{3}-3x^{2}-16x+48} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&0&-25&0&144\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 1 }&0&-25&0&144\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&0&-25&0&144\\& & \color{blue}{-3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-3&1&\color{orangered}{ 0 }&-25&0&144\\& & \color{orangered}{-3} & & & \\ \hline &1&\color{orangered}{-3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&0&-25&0&144\\& & -3& \color{blue}{9} & & \\ \hline &1&\color{blue}{-3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -25 } + \color{orangered}{ 9 } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrrr}-3&1&0&\color{orangered}{ -25 }&0&144\\& & -3& \color{orangered}{9} & & \\ \hline &1&-3&\color{orangered}{-16}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&0&-25&0&144\\& & -3& 9& \color{blue}{48} & \\ \hline &1&-3&\color{blue}{-16}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 48 } = \color{orangered}{ 48 } $
$$ \begin{array}{c|rrrrr}-3&1&0&-25&\color{orangered}{ 0 }&144\\& & -3& 9& \color{orangered}{48} & \\ \hline &1&-3&-16&\color{orangered}{48}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 48 } = \color{blue}{ -144 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&0&-25&0&144\\& & -3& 9& 48& \color{blue}{-144} \\ \hline &1&-3&-16&\color{blue}{48}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 144 } + \color{orangered}{ \left( -144 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-3&1&0&-25&0&\color{orangered}{ 144 }\\& & -3& 9& 48& \color{orangered}{-144} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{-16}&\color{blue}{48}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-3x^{2}-16x+48 } $ with a remainder of $ \color{red}{ 0 } $.