Answer
$\dfrac{3x^2+49}{x(x+7)^2}=\dfrac{1}{x}+\dfrac{2}{x+7}-\dfrac{28}{(x+7)^2}$
Work Step by Step
We are given the fraction:
$\dfrac{3x^2+49}{x(x+7)^2}$
As the denominator is already factored, we can write the partial fraction decomposition:
$\dfrac{3x^2+49}{x(x+7)^2}=\dfrac{A}{x}+\dfrac{B}{x+7}+\dfrac{C}{(x+7)^2}$
Multiply all terms by the least common denominator $x(x+7)^2$:
$x(x+7)^2\cdot\dfrac{3x^2+49}{x(x+7)^2}=x(x+7)^2\cdot\dfrac{A}{x}+x(x+7)^2\cdot\dfrac{B}{x+7}+x(x+7)^2\cdot\dfrac{C}{(x+7)^2}$
$3x^2+49=A(x+7)^2+Bx(x+7)+Cx$
$3x^2+49=Ax^2+14Ax+49A+Bx^2+7Bx+Cx$
$3x^2+49=(A+B)x^2+(14A+7B+C)x+49A$
Identify the coefficients and build the system of equations:
$\begin{cases}
A+B=3\\
14A+7B+C=0\\
49A=49
\end{cases}$
Solve the system:
$A=1$
$1+B=3$
$B=2$
$14(1)+7(2)+C=0$
$28+C=2$
$C=-28$
The partial fraction decomposition is:
$\dfrac{3x^2+49}{x(x+7)^2}=\dfrac{1}{x}+\dfrac{2}{x+7}-\dfrac{28}{(x+7)^2}$
