Divide using synthetic division $$ ( 12x^{4}-67x^{3}+268x^{2}+137x-50 ) \div ( 4x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&12&-67&268&137&-50\\& & -36& 309& -1731& \color{black}{4782} \\ \hline &\color{blue}{12}&\color{blue}{-103}&\color{blue}{577}&\color{blue}{-1594}&\color{orangered}{4732} \end{array} $$The solution is:
$$ \frac{ 12x^{4}-67x^{3}+268x^{2}+137x-50 }{ x+3 } = \color{blue}{12x^{3}-103x^{2}+577x-1594} ~+~ \frac{ \color{red}{ 4732 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&12&-67&268&137&-50\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 12 }&-67&268&137&-50\\& & & & & \\ \hline &\color{orangered}{12}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 12 } = \color{blue}{ -36 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&12&-67&268&137&-50\\& & \color{blue}{-36} & & & \\ \hline &\color{blue}{12}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -67 } + \color{orangered}{ \left( -36 \right) } = \color{orangered}{ -103 } $
$$ \begin{array}{c|rrrrr}-3&12&\color{orangered}{ -67 }&268&137&-50\\& & \color{orangered}{-36} & & & \\ \hline &12&\color{orangered}{-103}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -103 \right) } = \color{blue}{ 309 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&12&-67&268&137&-50\\& & -36& \color{blue}{309} & & \\ \hline &12&\color{blue}{-103}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 268 } + \color{orangered}{ 309 } = \color{orangered}{ 577 } $
$$ \begin{array}{c|rrrrr}-3&12&-67&\color{orangered}{ 268 }&137&-50\\& & -36& \color{orangered}{309} & & \\ \hline &12&-103&\color{orangered}{577}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 577 } = \color{blue}{ -1731 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&12&-67&268&137&-50\\& & -36& 309& \color{blue}{-1731} & \\ \hline &12&-103&\color{blue}{577}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 137 } + \color{orangered}{ \left( -1731 \right) } = \color{orangered}{ -1594 } $
$$ \begin{array}{c|rrrrr}-3&12&-67&268&\color{orangered}{ 137 }&-50\\& & -36& 309& \color{orangered}{-1731} & \\ \hline &12&-103&577&\color{orangered}{-1594}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -1594 \right) } = \color{blue}{ 4782 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&12&-67&268&137&-50\\& & -36& 309& -1731& \color{blue}{4782} \\ \hline &12&-103&577&\color{blue}{-1594}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -50 } + \color{orangered}{ 4782 } = \color{orangered}{ 4732 } $
$$ \begin{array}{c|rrrrr}-3&12&-67&268&137&\color{orangered}{ -50 }\\& & -36& 309& -1731& \color{orangered}{4782} \\ \hline &\color{blue}{12}&\color{blue}{-103}&\color{blue}{577}&\color{blue}{-1594}&\color{orangered}{4732} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 12x^{3}-103x^{2}+577x-1594 } $ with a remainder of $ \color{red}{ 4732 } $.