Partial fraction expansion

1. Sep 10, 2015

mathnovice

1. The problem statement, all variables and given/known data
divide in partial fractions:

x^3 -3x^2+x-12 / x^4+5x^2+4

2. Relevant equations

3. The attempt at a solution

I factored x^4+5x^2+4 to (x^2 +1) (x^2+4)

x^3 -3x^2+x-12/(x^2 +1) (x^2+4) = A (x^2+1)/(x^2 +1) (x^2+4) + B (x^2+4) /(x^2 +1) (x^2+4)

they all have the same denominator so: x^3-3x^2+x-12 = A (x^2+1) + B (x^2+4)

now normally my next step would be to try to solve for A by making ( x^2+4) equal to zero so I don't have B in my equation anymore. so I should find an x where x^2 is -4 but this is impossible because of the ^2. how do I solve this? thanks in advance

2. Sep 10, 2015

DEvens

You can't use just A and B. You need (ax+b) and (cx+d). You have an x^3 term.

3. Sep 10, 2015

Ray Vickson

You have written
$$x^3 - 3x^2 + x - \frac{12}{x^4} + 5 x^2 + 4$$
Are you sure that is what you really want?

If you mean
$$\frac{x^3 - 3 x^2 + x - 12}{x^4 + 5 x^2 + 4},$$
then you absolutely MUST use parentheses, like this: (x^3 -3x^2+x-12) / (x^4+5x^2+4).

If I were doing it I would expand $1/(x^4 + 5 x^2 + 4) = 1/(t^2 + 5 t + 4)$ into partial fractions in the variable $t =x^2$, then multiply by the numerator $x^3 - 3x^2 + x - 12$ later. The numerator can be written as
$$x^3 -3 x^2 + x - 12 = (x^3 + x) - (3x^2 + 3) - 9 = (x-3)(x^2+1) - 9$$
for the $(x^2+1)$ denominator and as
$$x^3 -3 x^2 + x - 12 = (x^3 + 4x) - (3x^2 + 12) - 3x = (x-3)(x^2 + 4) - 3x$$
for the $(x^2+ 4)$ denominator; that will allow a complete partial-fraction expansion.

The only remaining issue is whether the questioner wants/allows partial fractions in $x^2$, or whether you need to go all the way down to fractions in $x$. In the latter case your partial fractions will be of the form $1/(x + i c)$, where $c$ is a real number while $i = \sqrt{-1}$ is the imaginary unit.

Last edited: Sep 10, 2015
4. Sep 10, 2015

Staff: Mentor

No, this isn't right. You have skipped a step here, and your partial fraction decompositions aren't right. Try this to start with:
$\frac{x^3 -3x^2+x-12}{(x^2 +1) (x^2+4)} = \frac{Ax + B}{x^2 +1} + \frac{Cx + D}{x^2 +4}$

5. Sep 10, 2015

mathnovice

how do you guys right fractions and expontents that way? :D would come in handy.

Would this step be wrong?: (x^3-3x^2+x-12)/[(x^2+1)(x^2+4)] = [(Ax +B) (x^2+1)]/[(x^2 +1) (x^2+4)] + [(Cx+D)(x^2+4)] /[(x^2 +1) (x^2+4)]

and then say (x^3-3x^2+x-12) = (Ax +B) (x^2+1) + (Bx+c)(x^2+4) becose they all had the same denominator.

and if so then what is the next step? could turn it into (x^3-3x^2+x-12)= Ax^3 +Bx^2 +Ax +B+ Cx^3 +Dx^2 +4Cx +4D but I don't think that's really useful

thanks for helping me, was able to solve a lot of the exercises with watching the videos on khanacademy.com on partial fraction expansion but now i'm stuck

Last edited: Sep 10, 2015
6. Sep 10, 2015

DEvens

To see how somebody has done some equations, click the reply button and look at the tags. Just don't post it because it would be a duplication.

You wrote (Ax+B) and (Bx+C). You used B twice. You want (Ax+B) and (CX+D).

7. Sep 10, 2015

vela

Staff Emeritus
At this point, you can collect terms on the righthand side, giving you
$$x^3 - 3x^2 + x - 12 = (A+C)x^3 + (B+D)x^2 + (A+4C) x + (B+4D),$$ and then match coefficients. For example, from the $x^3$ terms, you'd get 1 = A+C. You'll have a system of four linear equations and four unknowns, which is straightforward to solve.

8. Sep 10, 2015

mathnovice

I found the answer: 1/(x^2+1) - 3/(x^2+4). everybody thanks for the help. I would never been able find the answer without it!

9. Sep 10, 2015

vela

Staff Emeritus
Unfortunately, that's not correct.