Divide using synthetic division $$ ( -7x^{3}-10 ) \div ( x-3 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&-7&0&0&-10\\& & -21& -63& \color{black}{-189} \\ \hline &\color{blue}{-7}&\color{blue}{-21}&\color{blue}{-63}&\color{orangered}{-199} \end{array} $$The solution is:
$$ \dfrac{ -7x^{3}-10 }{ x-3 } = \color{blue}{-7x^{2}-21x-63} \color{red}{~-~} \dfrac{ \color{red}{ 199 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-7&0&0&-10\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ -7 }&0&0&-10\\& & & & \\ \hline &\color{orangered}{-7}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -21 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-7&0&0&-10\\& & \color{blue}{-21} & & \\ \hline &\color{blue}{-7}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -21 \right) } = \color{orangered}{ -21 } $
$$ \begin{array}{c|rrrr}3&-7&\color{orangered}{ 0 }&0&-10\\& & \color{orangered}{-21} & & \\ \hline &-7&\color{orangered}{-21}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -21 \right) } = \color{blue}{ -63 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-7&0&0&-10\\& & -21& \color{blue}{-63} & \\ \hline &-7&\color{blue}{-21}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -63 \right) } = \color{orangered}{ -63 } $
$$ \begin{array}{c|rrrr}3&-7&0&\color{orangered}{ 0 }&-10\\& & -21& \color{orangered}{-63} & \\ \hline &-7&-21&\color{orangered}{-63}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -63 \right) } = \color{blue}{ -189 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-7&0&0&-10\\& & -21& -63& \color{blue}{-189} \\ \hline &-7&-21&\color{blue}{-63}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ \left( -189 \right) } = \color{orangered}{ -199 } $
$$ \begin{array}{c|rrrr}3&-7&0&0&\color{orangered}{ -10 }\\& & -21& -63& \color{orangered}{-189} \\ \hline &\color{blue}{-7}&\color{blue}{-21}&\color{blue}{-63}&\color{orangered}{-199} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -7x^{2}-21x-63 } $ with a remainder of $ \color{red}{ -199 } $.