How do you factor #x^4 + 4#?

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Answer 1 · 2015-05-15T23:30:03

#x^4+4# has no linear factors since #x^4+4 > 0# for all real values of #x#.

How about quadratic factors?

#x^4+4 = (x^2+ax+b)(x^2+cx+d)#

#= x^4+(a+c)x^3 + (b+d+ac)x^2 + (ad+bc)x + bd#

Comparing coefficients of #x^3# we must have #a+c=0#, so #c = -a#

... #=x^4+(b+d-a^2)x^2 + a(d-b)x + bd#

Looking at the coefficients of #x#, we either have #a = 0# or #b = d#.

If #a=0# then #b+d = 0# so #d=-b# and #bd=-b^2#, which would require #b^2 = -4# - not possible for real values of #b#.

If #b = d#, then since #bd = 4#, #b = d = 2#, which would make #a^2 = b+d = 4#, so #a=+-2#.

Indeed #x^4 + 4 = (x^2+2x+2)(x^2-2x+2)#