Divide using synthetic division $$ ( -3x^{4}+15x^{3}-2x+10 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&-3&15&0&-2&10\\& & -15& 0& 0& \color{black}{-10} \\ \hline &\color{blue}{-3}&\color{blue}{0}&\color{blue}{0}&\color{blue}{-2}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ -3x^{4}+15x^{3}-2x+10 }{ x-5 } = \color{blue}{-3x^{3}-2} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&-3&15&0&-2&10\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ -3 }&15&0&-2&10\\& & & & & \\ \hline &\color{orangered}{-3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&-3&15&0&-2&10\\& & \color{blue}{-15} & & & \\ \hline &\color{blue}{-3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}5&-3&\color{orangered}{ 15 }&0&-2&10\\& & \color{orangered}{-15} & & & \\ \hline &-3&\color{orangered}{0}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&-3&15&0&-2&10\\& & -15& \color{blue}{0} & & \\ \hline &-3&\color{blue}{0}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}5&-3&15&\color{orangered}{ 0 }&-2&10\\& & -15& \color{orangered}{0} & & \\ \hline &-3&0&\color{orangered}{0}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&-3&15&0&-2&10\\& & -15& 0& \color{blue}{0} & \\ \hline &-3&0&\color{blue}{0}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 0 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}5&-3&15&0&\color{orangered}{ -2 }&10\\& & -15& 0& \color{orangered}{0} & \\ \hline &-3&0&0&\color{orangered}{-2}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&-3&15&0&-2&10\\& & -15& 0& 0& \color{blue}{-10} \\ \hline &-3&0&0&\color{blue}{-2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}5&-3&15&0&-2&\color{orangered}{ 10 }\\& & -15& 0& 0& \color{orangered}{-10} \\ \hline &\color{blue}{-3}&\color{blue}{0}&\color{blue}{0}&\color{blue}{-2}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -3x^{3}-2 } $ with a remainder of $ \color{red}{ 0 } $.