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

2 Answers
Aug 8, 2015

Answer:

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

Answer:

#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.