Partial Fractions and power of a factor with $x^2$

Question

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.

Answer 1

You already have

\begin{align*} &\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}\\ &= \frac{(Ax + B)(x + 2) + (Cx + D)(x^2 + x + 1)(x + 2) + E(x^2 + x + 1)^2}{(x+2)(x^2+x+1)^2}. \end{align*}

Multiplying both sides by $(x+2)(x^2+x+1)^2$ we have $$3x^2 + 2x + 1 = (Ax + B)(x + 2) + (Cx + D)(x^2 + x + 1)(x + 2) + E(x^2 + x + 1)^2.$$ The right hand side is a quartic equation. If you expand all the brackets, you can collect all the like terms to get an equation as follows: $$3x^2 + 2x + 1 = k_4x^4 + k_3x^3+k_2x^2+k_1x+k_0$$ where $k_i$ depends on $A, B, C, D,$ and $E$. Then you can compare the coefficients of $1, x, x^2, x^3,$ and $x^4$ to get five equations in five unknowns which will be enough to determine $A, B, C, D,$ and $E$.

Answer 2

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.