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