How do I find the partial-fraction decomposition of (2x^3+7x^2-2x+6)/(x^4+4)?

2 Answers
Aug 8, 2015

First factor the denominator to find x^4+4 = (x^2+2x+2)(x^2-2x+2)

Then solve 2x^3+7x^2-2x+6 = (x^2-2x+2)(Ax+B) + (x^2+2x+2)(Cx+D)

finding A=0, B=3, C=2 and D=0.

Explanation:

x^4+4 has no linear factors with Real coefficients since x^4+4 >= 4 > 0 for all x in RR.

It will have quadratic factors with Real coefficients.

x^4+4 = (ax^2+bx+c)(dx^2+ex+f)

Without loss of generalisation, a = d = 1, so

x^4+4 = (x^2+bx+c)(x^2+ex+f)

Then looking at the coefficient of x^3, we have e = -b, so

x^4+4 = (x^2+bx+c)(x^2-bx+f)

Then looking at the coefficient of x we have b(f-c) = 0, so either b=0 or c = f.

If b=0 then c+f = 0 and cf = -c^2 = 4, which has no Real solutions (it has solutions c = 2i, f = -2i).

If c = f and cf = 4 then c=f=+-2. Then the coefficient of x^2 tells us f+c-b^2 = 0. Now if b is Real, then b^2 >= 0, so f+c >= 0, hence f=c=2. Then b^2 = 4, so b = +-2.

So we have x^4+4 = (x^2+2x+2)(x^2-2x+2)

Now attempt to solve:

(2x^3+7x^2-2x+6)/(x^4+4) = (Ax+B)/(x^2+2x+2) + (Cx+D)/(x^2-2x+2)

If we multiply through by (x^4+4) = (x^2+2x+2)(x^2-2x+2) then this becomes:

2x^3+7x^2-2x+6

= (x^2-2x+2)(Ax+B) + (x^2+2x+2)(Cx+D)

= (A+C)x^3+(B-2A+D+2C)x^2+2(A-B+C+D)x+2(B+D)

Equating coefficients we get:

(i) A+C = 2
(ii) B-2A+D+2C = 7
(iii) 2(A-B+C+D) = -2
(iv) 2(B+D) = 6

From (i) and (iv) we get:

(v) C = 2 - A and
(vi) D=3-B

Substitute these into (ii) to get:

color(red)(cancel(color(black)(B)))-2A+3-color(red)(cancel(color(black)(B)))+4-2A=7

Hence A=0. Then from (v) C = 2.

Substitute A=0, C=2 and D=3-B into (iii) to get:

2(0-B+2+3-B) = -2

Divide both sides by 2 to find:

5-2B = -1

Hence B=3. Then from (vi) D=0.

So

(2x^3+7x^2-2x+6)/(x^4+4) = 3/(x^2+2x+2) + (2x)/(x^2-2x+2)

Aug 9, 2015

x^4+4 = (x^2+2x+2)(x^2-2x+2) hence

(2x^3+7x^2-2x+6)/(x^4+4) = 3/(x^2+2x+2) + (2x)/(x^2-2x+2)

Explanation:

Here is a quick way to find the quadratic factors of x^4+4 using complex arithmetic ...

sqrt(-4) = 2i

sqrt(2i) = i + 1

If x_1 is a root of x^4+4 = 0, then so are -x_1, bar(x_1) and bar(-x_1), where bar(x_1) means the complex conjugate. This four-fold symmetry occurs due to the only power of x being x^4.

Here are the roots in the complex plane:

graph{((x-1)^2+(y-1)^2 - 0.01)((x+1)^2+(y-1)^2 - 0.01)((x-1)^2+(y+1)^2 - 0.01)((x+1)^2+(y+1)^2-0.01) = 0 [-5, 5, -2.5, 2.5]}

So x^4+4 = (x-1-i)(x-1+i)(x+1-i)(x+1+i)

Now pick pairs of factors to multiply to get Real coefficients.

To do this, pick the ones which are complex conjugates:

(x+1-i)(x+1+i) = ((x+1)-i)((x+1)+i)

= (x+1)^2-i^2 = x^2+2x+1+1 = x^2+2x+2

and

(x-1-i)(x-1+i) = ((x-1)-i)((x-1)+i)

= (x-1)^2-i^2 = x^2-2x+1+1 = x^2-2x+2

For the rest of the partial fraction decomposition, see the other answers.