Answer
$\dfrac{4x^3+5x^2+7x-1}{(x^2+x+1)^2}=\dfrac{4x+1}{x^2+x+1}+\dfrac{2x-2}{(x^2+x+1)^2}$
Work Step by Step
We are given the fraction:
$\dfrac{4x^3+5x^2+7x-1}{(x^2+x+1)^2}$
As the denominator is already factored, we can write the partial fraction decomposition:
$\dfrac{4x^3+5x^2+7x-1}{(x^2+x+1)^2}=\dfrac{Ax+B}{x^2+x+1}+\dfrac{Cx+D}{(x^2+x+1)^2}$
Multiply all terms by the least common denominator $(x^2+x+1)^2$:
$(x^2+x+1)^2\cdot\dfrac{4x^3+5x^2+7x-1}{(x^2+x+1)^2}=(x^2+x+1)^2\cdot\dfrac{Ax+B}{x^2+x+1}+(x^2+x+1)^2\cdot\dfrac{Cx+D}{(x^2+x+1)^2}$
$4x^3+5x^2+7x-1=(Ax+B)(x^2+x+1)+(Cx+D)$
$4x^3+5x^2+7x-1=Ax^3+Ax^2+Ax+Bx^2+Bx+B+Cx+D$
$4x^3+5x^2+7x-1=Ax^3+(A+B)x^2+(A+B+C)x+(B+D)$
Identify the coefficients and build the system of equations:
$\begin{cases}
A=4\\
A+B=5\\
A+B+C=7\\
B+D=-1
\end{cases}$
Solve the system:
$A=4$
$A+B=5$
$4+B=5$
$B=1$
$A+B+C=7$
$4+1+C=7$
$5+C=7$
$C=2$
$B+D=-1$
$1+D=-1$
$D=-2$
The partial fraction decomposition is:
$\dfrac{4x^3+5x^2+7x-1}{(x^2+x+1)^2}=\dfrac{4x+1}{x^2+x+1}+\dfrac{2x-2}{(x^2+x+1)^2}$
