<s>[INST] Partial Fractions and power of a factor with $x^2$ I just started working with partial fractions and hit a wall with splitting this one:
$$ \frac{3x^2 + 2x + 1}{(x + 2)(x^2 + x + 1)^2} $$
I get here:
$$ \frac{Ax + B}{(x^2 + x + 1)^2} + \frac{Cx + D}{x^2 + x + 1} + \frac{E}{x + 2}$$
Then on to:
$$ (Ax + B)(x + 2) + (Cx + D)(x^2 + x + 1)(x + 2) + E(x^2 + x + 1)^2 $$
I find that $E = 1$ by using $x = -2$. I am unsure how to proceed from here.
 [/INST] You don't have any equations there.
Presumably, you start with
$$\frac{3x^2 + 2x + 1}{(x + 2)(x^2 + x + 1)^2} 
= \frac{Ax + B}{(x^2 + x + 1)^2} + \frac{Cx + D}{x^2 + x + 1} + \frac{E}{x + 2}
$$
then multiply both sides
by $ (x + 2)(x^2 + x + 1)^2$
to get
$$3x^2 + 2x + 1
=(Ax + B)(x + 2) + (Cx + D)(x^2 + x + 1)(x + 2) + E(x^2 + x + 1)^2
.$$
There are a variety of ways
to go from here.
As you did,
when you set
$x = -2$,
you get
$12-4+1=9E$
so $E = 1$.
The equation then becomes
$$3x^2 + 2x + 1
=(Ax + B)(x + 2) + (Cx + D)(x^2 + x + 1)(x + 2) + (x^2 + x + 1)^2
.$$
You can do as
Michael Albanese
suggested to get equations for the
other coefficients.
This is probably the
best way.
You can look at the roots
of
$x^2 + x + 1=0$,
which are complex.
You can let $x=0$,
and this becomes
$1
=2B + 2D + 1
$,
or $B = -D$.
Looking at the coefficient of $x^4$,
you get
$Cx^4+x^4 = 0$
or $C = -1$.
I'll stop here.
</s>