Answer
The simplified partial fraction expansion is $\frac{3}{x-2}+\frac{x-1}{{{x}^{2}}+2x+4}$.
Work Step by Step
The provided rational fraction expression is as follows:
$\frac{4{{x}^{2}}+3x+14}{{{x}^{3}}-8}$
Then, demonstrating the steps as given below:
$\begin{align}
& \frac{4{{x}^{2}}+3x+14}{{{x}^{3}}-8}=\frac{4{{x}^{2}}+3x+14}{{{x}^{3}}-{{2}^{3}}} \\
& =\frac{4{{x}^{2}}+3x+14}{\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)}
\end{align}$
Step 1:
Set up the partial fraction expansion with unknown constant coefficients and then write a constant coefficient over each of the two distinct algebraic linear factors in the denominator of the expression.
$\frac{4{{x}^{2}}+3x+14}{\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)}=\frac{A}{x-2}+\frac{Bx+C}{{{x}^{2}}+2x+4}$
Step 2:
Multiply both sides of the equation by the expression $\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)$, considering the least common denominator.
$\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)\times \left( \frac{4{{x}^{2}}+3x+14}{\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)} \right)=\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)\times \left( \frac{A}{x-2}+\frac{Bx+C}{{{x}^{2}}+2x+4} \right)$
Then, solving both sides of the equation, we get:
$4{{x}^{2}}+3x+14=\left( {{x}^{2}}+2x+4 \right)A+\left( Bx+C \right)\left( x-2 \right)$
$\begin{align}
& 4{{x}^{2}}+3x+14=\left( {{x}^{2}}+2x+4 \right)A+\left( Bx+C \right)\left( x-2 \right) \\
& =A\left( {{x}^{2}}+2x+4 \right)+B\left( {{x}^{2}}-2x \right)+C\left( x-2 \right) \\
& ={{x}^{2}}\left( A+B \right)+x\left( 2A-2B+C \right)+4A-2C
\end{align}$
Step 3:
Then equating the coefficients of like powers of $ x $ and of the constant terms, we get the system of linear equations with the unknowns values of $ A $, $ B $, $ C $ and $ D $.
$ A+B=4$ (I)
$2A-2B+C=3$ (II)
$4A-2C=14$ (III)
Step 4:
Solve the system for $ A $, $ B $, $ C $ and $ D $.
Eliminate $ C $ from the equation by multiplying by 2 in equation (II) and then adding equations (II) and (III) as given below:
$\begin{align}
& \left( 4A-4B+2C \right)+4A-2C=6+14 \\
& 8A-4B=20
\end{align}$
$2A-B=5$ (IV)
Also, eliminate $ B $ from the equation by adding equations (I) and (IV) as given below:
$\begin{align}
& \left( A+B \right)+2A-B=4+5 \\
& 3A=9 \\
& A=3
\end{align}$
Substitute the values of $ A $ in equation (I) and simplify as given below:
$\begin{align}
& A+B=4 \\
& 3+B=4 \\
& B=4-3 \\
& B=1
\end{align}$
Similarly, put the values of $ A $ and $ B $ in equation (II) and find the value of $ C $ as given below:
$\begin{align}
& 2A-2B+C=3 \\
& 2\times 3-2\times 1+C=3 \\
& C=3+2-6 \\
& C=-1
\end{align}$
Step 5:
Replace the values of $ A $, $ B $, and, $ C $, writing the partial function expression as given below:
$\begin{align}
& \frac{4{{x}^{2}}+3x+14}{\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)}=\frac{A}{x-2}+\frac{Bx+C}{{{x}^{2}}+2x+4} \\
& =\frac{3}{x-2}+\frac{1\times x-1}{{{x}^{2}}+2x+4} \\
& =\frac{3}{x-2}+\frac{x-1}{{{x}^{2}}+2x+4}
\end{align}$
Thus, $\frac{3}{x-2}+\frac{x-1}{{{x}^{2}}+2x+4}$ is the required partial fraction expansion of $\frac{4{{x}^{2}}+3x+14}{{{x}^{3}}-8}$.
