# Split into partial fractions $\frac{3x^2+2x+1}{(x+2)(x^2+x+1)^2}$

This is from "Calculus Made Easy", Exercises 10, Question 15 (page 147). I've worked this one over and over and still haven't made progress.

This is my initial setup:

$$\frac{3x^2+2x+1}{(x+2)(x^2+x+1)^2} = \frac{A}{(x+2)} + \frac{Bx+C}{(x^2+x+1)} + \frac{Dx+E}{(x^2+x+1)^2}$$

then:

$$3x^2+2x+1 = A(x^2+x+1)^2 + (Bx+C)(x+2)(x^2+x+1) + (Dx+E)(x+2)$$

and I can solve for A by setting $$x=-2$$, which yields $$A=1$$.

I know the final answer is this:

$$\frac{1}{x+2} - \frac{x-1}{x^2+x+1} - \frac{1}{(x^2+x+1)^2}$$

But I've worked this problem many ways and cannot make progress on the numerators for the other fractions.

• There's an $x^4$ on the RHS that can only be cancelled by the $A,B$ coefficients. What relationship between them accomplishes this? Commented Dec 2, 2019 at 2:12
• Consider $3x^2+2x+1 \equiv A(x^2+x+1)^2 + (Bx+C)(x+2)(x^2+x+1) + (Dx+E)(x+2)$ and equate coefficients for $x^k$ in either side, you'll get system of $5$ linear equations. Or also try "setting $x$" to [the roots of $x^2+x+1=0$] $-\frac12\pm\frac{\sqrt{3}}2i$ ) Commented Dec 2, 2019 at 2:13
• @abiessu: I'm not seeing that. I have a $x^4$ on the RHS and no $x^4$ term on the left, so $x^4=0$?
– Mike
Commented Dec 2, 2019 at 3:06
• @Mike: no, the $A,B$ coefficients add to make $0x^4$ in the final result, but no specific value of $x$ is necessary... Commented Dec 2, 2019 at 5:50

$$\frac{3x^2+2x+1}{(x+2)(x^2+x+1)^2}= \frac {(3x^2+3x +3)-(x+2)}{(x+2)(x^2+x+1)^2}=$$

$$\frac {3}{(x+2)(x^2+x+1)}-\frac { 1}{(x^2+x+1)^2}$$

Now you may proceed with the first fraction.

$$\frac {3}{(x+2)(x^2+x+1)}=\frac {1}{x+2} -\frac {x-1}{x^2+x+1}$$

• I like this one. I had not considered pulling $(x+2)$ out of the numerator.
– Mike
Commented Dec 2, 2019 at 2:52

I'll expand all the things from the comment.
Method 1:
By expanding we get $$\begin{cases} A+B=0&&\hbox{ for }x^4\\\ 2A+3B+C=0&&\hbox{ for }x^3\\\ 3 A + 3 B + 3 C + D - 3=0&&\hbox{ for }x^2\\\ 2 A + 2 B + 3 C + 2 D + E - 2=0&&\hbox{ for }x\\\ A + 2 C + 2 E - 1=0&&\hbox{ for }1 \end{cases}$$ which boils down to $$A = 1, B = -1, C = 1, D = 0, E = -1$$.
Method 2 "substitution":
with $$x=-\frac12\pm\frac{\sqrt{3}}{2}i$$ we get rid of $$(x^2+x+1)$$, can have equations for $$D,E$$ and after finding $$D,E$$ and bringing them to the LHS we can cancel by $$(x+2)$$ which boils down to Mohammad Riazi-Kermani's answer

• This is good. I always forget about creating a set of equations.
– Mike
Commented Dec 3, 2019 at 1:43

Now that you know $$A=1$$ set $$A$$ equal to $$1$$. You get

$$3x^2+2x+1 = (x^2+x+1)^2 + (Bx+C)(x+2)(x^2+x+1) + (Dx+E)(x+2)$$ $$3x^2+2x+1 - (x^4+2x^3+3x^2+2x+1) = (Bx+C)(x+2)(x^2+x+1) + (Dx+E)(x+2)$$ $$-x^4-2x^3 = (Bx+C)(x+2)(x^2+x+1) + (Dx+E)(x+2)$$ $$-x^3(x+2) = (Bx+C)(x+2)(x^2+x+1) + (Dx+E)(x+2)$$ $$-x^3 = (Bx+C)(x^2+x+1) + (Dx+E)$$ $$-x^3 = Bx^3 + (B+C)x^2 + (B+C+D)x + (C+E)$$

and proceed from there

• Maybe $x^3$ in the later equations on the LHS? Commented Dec 2, 2019 at 2:14