Divide using synthetic division $$ ( 3x^{4}+7x^{3}-15x^{2}+x-2 ) \div ( x+7 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-7&3&7&-15&1&-2\\& & -21& 98& -581& \color{black}{4060} \\ \hline &\color{blue}{3}&\color{blue}{-14}&\color{blue}{83}&\color{blue}{-580}&\color{orangered}{4058} \end{array} $$The solution is:
$$ \frac{ 3x^{4}+7x^{3}-15x^{2}+x-2 }{ x+7 } = \color{blue}{3x^{3}-14x^{2}+83x-580} ~+~ \frac{ \color{red}{ 4058 } }{ x+7 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-7}&3&7&-15&1&-2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-7&\color{orangered}{ 3 }&7&-15&1&-2\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -7 } \cdot \color{blue}{ 3 } = \color{blue}{ -21 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-7}&3&7&-15&1&-2\\& & \color{blue}{-21} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -21 \right) } = \color{orangered}{ -14 } $
$$ \begin{array}{c|rrrrr}-7&3&\color{orangered}{ 7 }&-15&1&-2\\& & \color{orangered}{-21} & & & \\ \hline &3&\color{orangered}{-14}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -7 } \cdot \color{blue}{ \left( -14 \right) } = \color{blue}{ 98 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-7}&3&7&-15&1&-2\\& & -21& \color{blue}{98} & & \\ \hline &3&\color{blue}{-14}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 98 } = \color{orangered}{ 83 } $
$$ \begin{array}{c|rrrrr}-7&3&7&\color{orangered}{ -15 }&1&-2\\& & -21& \color{orangered}{98} & & \\ \hline &3&-14&\color{orangered}{83}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -7 } \cdot \color{blue}{ 83 } = \color{blue}{ -581 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-7}&3&7&-15&1&-2\\& & -21& 98& \color{blue}{-581} & \\ \hline &3&-14&\color{blue}{83}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -581 \right) } = \color{orangered}{ -580 } $
$$ \begin{array}{c|rrrrr}-7&3&7&-15&\color{orangered}{ 1 }&-2\\& & -21& 98& \color{orangered}{-581} & \\ \hline &3&-14&83&\color{orangered}{-580}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -7 } \cdot \color{blue}{ \left( -580 \right) } = \color{blue}{ 4060 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-7}&3&7&-15&1&-2\\& & -21& 98& -581& \color{blue}{4060} \\ \hline &3&-14&83&\color{blue}{-580}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 4060 } = \color{orangered}{ 4058 } $
$$ \begin{array}{c|rrrrr}-7&3&7&-15&1&\color{orangered}{ -2 }\\& & -21& 98& -581& \color{orangered}{4060} \\ \hline &\color{blue}{3}&\color{blue}{-14}&\color{blue}{83}&\color{blue}{-580}&\color{orangered}{4058} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}-14x^{2}+83x-580 } $ with a remainder of $ \color{red}{ 4058 } $.