Divide using synthetic division $$ ( 2x^{4}+10x^{3}-40x^{2}+61x-24 ) \div ( x+8 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-8&2&10&-40&61&-24\\& & -16& 48& -64& \color{black}{24} \\ \hline &\color{blue}{2}&\color{blue}{-6}&\color{blue}{8}&\color{blue}{-3}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+10x^{3}-40x^{2}+61x-24 }{ x+8 } = \color{blue}{2x^{3}-6x^{2}+8x-3} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 8 = 0 $ ( $ x = \color{blue}{ -8 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-8}&2&10&-40&61&-24\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-8&\color{orangered}{ 2 }&10&-40&61&-24\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ 2 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-8}&2&10&-40&61&-24\\& & \color{blue}{-16} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}-8&2&\color{orangered}{ 10 }&-40&61&-24\\& & \color{orangered}{-16} & & & \\ \hline &2&\color{orangered}{-6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-8}&2&10&-40&61&-24\\& & -16& \color{blue}{48} & & \\ \hline &2&\color{blue}{-6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -40 } + \color{orangered}{ 48 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}-8&2&10&\color{orangered}{ -40 }&61&-24\\& & -16& \color{orangered}{48} & & \\ \hline &2&-6&\color{orangered}{8}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ 8 } = \color{blue}{ -64 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-8}&2&10&-40&61&-24\\& & -16& 48& \color{blue}{-64} & \\ \hline &2&-6&\color{blue}{8}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 61 } + \color{orangered}{ \left( -64 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-8&2&10&-40&\color{orangered}{ 61 }&-24\\& & -16& 48& \color{orangered}{-64} & \\ \hline &2&-6&8&\color{orangered}{-3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-8}&2&10&-40&61&-24\\& & -16& 48& -64& \color{blue}{24} \\ \hline &2&-6&8&\color{blue}{-3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -24 } + \color{orangered}{ 24 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-8&2&10&-40&61&\color{orangered}{ -24 }\\& & -16& 48& -64& \color{orangered}{24} \\ \hline &\color{blue}{2}&\color{blue}{-6}&\color{blue}{8}&\color{blue}{-3}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-6x^{2}+8x-3 } $ with a remainder of $ \color{red}{ 0 } $.