Divide using synthetic division $$ ( 4x^{3}-6x^{2}-5x+8 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&4&-6&-5&8\\& & -12& 54& \color{black}{-147} \\ \hline &\color{blue}{4}&\color{blue}{-18}&\color{blue}{49}&\color{orangered}{-139} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-6x^{2}-5x+8 }{ x+3 } = \color{blue}{4x^{2}-18x+49} \color{red}{~-~} \frac{ \color{red}{ 139 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&-6&-5&8\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 4 }&-6&-5&8\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&-6&-5&8\\& & \color{blue}{-12} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -18 } $
$$ \begin{array}{c|rrrr}-3&4&\color{orangered}{ -6 }&-5&8\\& & \color{orangered}{-12} & & \\ \hline &4&\color{orangered}{-18}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -18 \right) } = \color{blue}{ 54 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&-6&-5&8\\& & -12& \color{blue}{54} & \\ \hline &4&\color{blue}{-18}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 54 } = \color{orangered}{ 49 } $
$$ \begin{array}{c|rrrr}-3&4&-6&\color{orangered}{ -5 }&8\\& & -12& \color{orangered}{54} & \\ \hline &4&-18&\color{orangered}{49}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 49 } = \color{blue}{ -147 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&-6&-5&8\\& & -12& 54& \color{blue}{-147} \\ \hline &4&-18&\color{blue}{49}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -147 \right) } = \color{orangered}{ -139 } $
$$ \begin{array}{c|rrrr}-3&4&-6&-5&\color{orangered}{ 8 }\\& & -12& 54& \color{orangered}{-147} \\ \hline &\color{blue}{4}&\color{blue}{-18}&\color{blue}{49}&\color{orangered}{-139} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-18x+49 } $ with a remainder of $ \color{red}{ -139 } $.