Divide using synthetic division $$ ( 4x^{3}-2x^{2}+4x-5 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&4&-2&4&-5\\& & -12& 42& \color{black}{-138} \\ \hline &\color{blue}{4}&\color{blue}{-14}&\color{blue}{46}&\color{orangered}{-143} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-2x^{2}+4x-5 }{ x+3 } = \color{blue}{4x^{2}-14x+46} \color{red}{~-~} \frac{ \color{red}{ 143 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&-2&4&-5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 4 }&-2&4&-5\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&-2&4&-5\\& & \color{blue}{-12} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -14 } $
$$ \begin{array}{c|rrrr}-3&4&\color{orangered}{ -2 }&4&-5\\& & \color{orangered}{-12} & & \\ \hline &4&\color{orangered}{-14}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -14 \right) } = \color{blue}{ 42 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&-2&4&-5\\& & -12& \color{blue}{42} & \\ \hline &4&\color{blue}{-14}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 42 } = \color{orangered}{ 46 } $
$$ \begin{array}{c|rrrr}-3&4&-2&\color{orangered}{ 4 }&-5\\& & -12& \color{orangered}{42} & \\ \hline &4&-14&\color{orangered}{46}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 46 } = \color{blue}{ -138 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&-2&4&-5\\& & -12& 42& \color{blue}{-138} \\ \hline &4&-14&\color{blue}{46}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -138 \right) } = \color{orangered}{ -143 } $
$$ \begin{array}{c|rrrr}-3&4&-2&4&\color{orangered}{ -5 }\\& & -12& 42& \color{orangered}{-138} \\ \hline &\color{blue}{4}&\color{blue}{-14}&\color{blue}{46}&\color{orangered}{-143} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-14x+46 } $ with a remainder of $ \color{red}{ -143 } $.