Divide using synthetic division $$ ( 2x^{3}+23x^{2}-26x+3 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&2&23&-26&3\\& & -10& -65& \color{black}{455} \\ \hline &\color{blue}{2}&\color{blue}{13}&\color{blue}{-91}&\color{orangered}{458} \end{array} $$The solution is:
$$ \frac{ 2x^{3}+23x^{2}-26x+3 }{ x+5 } = \color{blue}{2x^{2}+13x-91} ~+~ \frac{ \color{red}{ 458 } }{ x+5 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&2&23&-26&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 2 }&23&-26&3\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 2 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&2&23&-26&3\\& & \color{blue}{-10} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 23 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrr}-5&2&\color{orangered}{ 23 }&-26&3\\& & \color{orangered}{-10} & & \\ \hline &2&\color{orangered}{13}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 13 } = \color{blue}{ -65 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&2&23&-26&3\\& & -10& \color{blue}{-65} & \\ \hline &2&\color{blue}{13}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -26 } + \color{orangered}{ \left( -65 \right) } = \color{orangered}{ -91 } $
$$ \begin{array}{c|rrrr}-5&2&23&\color{orangered}{ -26 }&3\\& & -10& \color{orangered}{-65} & \\ \hline &2&13&\color{orangered}{-91}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -91 \right) } = \color{blue}{ 455 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&2&23&-26&3\\& & -10& -65& \color{blue}{455} \\ \hline &2&13&\color{blue}{-91}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 455 } = \color{orangered}{ 458 } $
$$ \begin{array}{c|rrrr}-5&2&23&-26&\color{orangered}{ 3 }\\& & -10& -65& \color{orangered}{455} \\ \hline &\color{blue}{2}&\color{blue}{13}&\color{blue}{-91}&\color{orangered}{458} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}+13x-91 } $ with a remainder of $ \color{red}{ 458 } $.