Answer
$\dfrac{5x^2+6x+3}{(x+1)(x^2+x+2)}=\dfrac{2}{x+1}+\dfrac{3x-1}{x^2+x+2}$
Work Step by Step
We are given the fraction:
$\dfrac{5x^2+6x+3}{(x+1)(x^2+x+2)}$
As the denominator is already factored, we can write the partial fraction decomposition:
$\dfrac{5x^2+6x+3}{(x+1)(x^2+x+2)}=\dfrac{A}{x+1}+\dfrac{Bx+C}{x^2+x+2}$
Multiply all terms by the least common denominator $(x+1)(x^2+x+2)$:
$(x+1)(x^2+x+2)\cdot\dfrac{5x^2+6x+3}{(x+1)(x^2+x+2)}=(x+1)(x^2+x+2)\cdot\dfrac{A}{x+1}+(x+1)(x^2+x+2)\cdot\dfrac{Bx+C}{x^2+x+2}$
$5x^2+6x+3=A(x^2+2x+2)+(Bx+C)(x+1)$
$5x^2+6x+3=Ax^2+2Ax+2A+Bx^2+Bx+Cx+C$
$5x^2+6x+3=(A+B)x^2+(2A+B+C)x+(2A+C)$
Identify the coefficients and build the system of equations:
$\begin{cases}
A+B=5\\
2A+B+C=6\\
2A+C=3
\end{cases}$
Solve the system: multiply Equation 3 by -1 and add it to Equation 2:
$\begin{cases}
A+B=5\\
2A+B+C-2A-C=6-3
\end{cases}$
$\begin{cases}
A+B=5\\
B=3
\end{cases}$
$A+B=5$
$A+3=5$
$A=2$
$2A+C=3$
$2(2)+C=3$
$C=-1$
The partial fraction decomposition is:
$\dfrac{5x^2+6x+3}{(x+1)(x^2+x+2)}=\dfrac{2}{x+1}+\dfrac{3x-1}{x^2+x+2}$
