Divide using synthetic division $$ ( 2x^{4}-5 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&2&0&0&0&-5\\& & -6& 18& -54& \color{black}{162} \\ \hline &\color{blue}{2}&\color{blue}{-6}&\color{blue}{18}&\color{blue}{-54}&\color{orangered}{157} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-5 }{ x+3 } = \color{blue}{2x^{3}-6x^{2}+18x-54} ~+~ \frac{ \color{red}{ 157 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&0&0&0&-5\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 2 }&0&0&0&-5\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 2 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&0&0&0&-5\\& & \color{blue}{-6} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}-3&2&\color{orangered}{ 0 }&0&0&-5\\& & \color{orangered}{-6} & & & \\ \hline &2&\color{orangered}{-6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&0&0&0&-5\\& & -6& \color{blue}{18} & & \\ \hline &2&\color{blue}{-6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 18 } = \color{orangered}{ 18 } $
$$ \begin{array}{c|rrrrr}-3&2&0&\color{orangered}{ 0 }&0&-5\\& & -6& \color{orangered}{18} & & \\ \hline &2&-6&\color{orangered}{18}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 18 } = \color{blue}{ -54 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&0&0&0&-5\\& & -6& 18& \color{blue}{-54} & \\ \hline &2&-6&\color{blue}{18}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -54 \right) } = \color{orangered}{ -54 } $
$$ \begin{array}{c|rrrrr}-3&2&0&0&\color{orangered}{ 0 }&-5\\& & -6& 18& \color{orangered}{-54} & \\ \hline &2&-6&18&\color{orangered}{-54}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -54 \right) } = \color{blue}{ 162 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&0&0&0&-5\\& & -6& 18& -54& \color{blue}{162} \\ \hline &2&-6&18&\color{blue}{-54}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 162 } = \color{orangered}{ 157 } $
$$ \begin{array}{c|rrrrr}-3&2&0&0&0&\color{orangered}{ -5 }\\& & -6& 18& -54& \color{orangered}{162} \\ \hline &\color{blue}{2}&\color{blue}{-6}&\color{blue}{18}&\color{blue}{-54}&\color{orangered}{157} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-6x^{2}+18x-54 } $ with a remainder of $ \color{red}{ 157 } $.