Answer
$\dfrac{4x^2+3x+14}{(x-2)(x^2+2x+4)}=\dfrac{3}{x-2}+\dfrac{x-1}{x^2+2x+4}$
Work Step by Step
We are given the fraction:
$\dfrac{4x^2+3x+14}{x^3-8}$
Factor the denominator:
$\dfrac{4x^2+3x+14}{x^3-8}=\dfrac{4x^2+3x+14}{(x-2)(x^2+2x+4)}$
We can write the partial fraction decomposition:
$\dfrac{4x^2+3x+14}{(x-2)(x^2+2x+4)}=\dfrac{A}{x-2}+\dfrac{Bx+C}{x^2+2x+4}$
Multiply all terms by the least common denominator $(x-2)(x^2+2x+4)$:
$(x-2)(x^2+2x+4)\cdot\dfrac{4x^2+3x+14}{(x-2)(x^2+2x+4)}=(x-2)(x^2+2x+4)\cdot\dfrac{A}{x-2}+(x-2)(x^2+2x+4)\cdot\dfrac{Bx+C}{x^2+2x+4}$
$4x^2+3x+14=A(x^2+2x+4)+(Bx+C)(x-2)$
$4x^2+3x+14=Ax^2+2Ax+4A+Bx^2-2Bx+Cx-2C$
$4x^2+3x+14=(A+B)x^2+(2A-2B+C)x+(4A-2C)$
Identify the coefficients and build the system of equations:
$\begin{cases}
A+B=4\\
2A-2B+C=3\\
4A-2C=14
\end{cases}$
$\begin{cases}
A+B=4\\
2A-2B+C=3\\
2A-C=7
\end{cases}$
Solve the system: Add Equation 2 to Equation 3:
$\begin{cases}
A+B=4\\
2A-2B+C+2A-C=3+7
\end{cases}$
$\begin{cases}
A+B=4\\
4A-2B=10
\end{cases}$
$\begin{cases}
A+B=4\\
2A-B=5
\end{cases}$
$A+B+2A-B=4+5$
$3A=9$
$A=3$
$A+B=4$
$3+B=4$
$B=1$
$2A-C=7$
$2(3)-C=7$
$6-C=7$
$C=-1$
The partial fraction decomposition is:
$\dfrac{4x^2+3x+14}{(x-2)(x^2+2x+4)}=\dfrac{3}{x-2}+\dfrac{x-1}{x^2+2x+4}$
