How do I Solve for A in Partial Fraction Expansion?

[mathnovice]

Homework Statement

divide in partial fractions:

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

Homework Equations

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

 

[DEvens]

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

 

[Ray Vickson]

Homework Statement

divide in partial fractions:

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

Homework Equations

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

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.

 

[Mark44]

Homework Statement

divide in partial fractions:

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

You need parentheses here -- one pair around the whole numerator and another pair around the whole denominator.

Homework Equations

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)

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

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

 

[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

 

[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).

 

[vela]

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

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.

 

[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!

 

[vela]

Unfortunately, that's not correct.