Divide using synthetic division $$ ( 4x^{4}+13x^{3}+8x^{2}+10x-15 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&4&13&8&10&-15\\& & -12& -3& -15& \color{black}{15} \\ \hline &\color{blue}{4}&\color{blue}{1}&\color{blue}{5}&\color{blue}{-5}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 4x^{4}+13x^{3}+8x^{2}+10x-15 }{ x+3 } = \color{blue}{4x^{3}+x^{2}+5x-5} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&13&8&10&-15\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 4 }&13&8&10&-15\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 4 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&13&8&10&-15\\& & \color{blue}{-12} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}-3&4&\color{orangered}{ 13 }&8&10&-15\\& & \color{orangered}{-12} & & & \\ \hline &4&\color{orangered}{1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&13&8&10&-15\\& & -12& \color{blue}{-3} & & \\ \hline &4&\color{blue}{1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}-3&4&13&\color{orangered}{ 8 }&10&-15\\& & -12& \color{orangered}{-3} & & \\ \hline &4&1&\color{orangered}{5}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 5 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&13&8&10&-15\\& & -12& -3& \color{blue}{-15} & \\ \hline &4&1&\color{blue}{5}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}-3&4&13&8&\color{orangered}{ 10 }&-15\\& & -12& -3& \color{orangered}{-15} & \\ \hline &4&1&5&\color{orangered}{-5}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&13&8&10&-15\\& & -12& -3& -15& \color{blue}{15} \\ \hline &4&1&5&\color{blue}{-5}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 15 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-3&4&13&8&10&\color{orangered}{ -15 }\\& & -12& -3& -15& \color{orangered}{15} \\ \hline &\color{blue}{4}&\color{blue}{1}&\color{blue}{5}&\color{blue}{-5}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}+x^{2}+5x-5 } $ with a remainder of $ \color{red}{ 0 } $.