How do you factor #x^4 + 3x^2 + 4#?

1 Answer
Aug 28, 2015

Answer:

Determine that #x^4+3x^2+4# only has quadratic factors with Real coefficients and solve some simultaneous equations to find:

#x^4+3x^2+4 = (x^2+x+2)(x^2-x+2)#

Explanation:

#x^4+3x^2+4# has no linear factors with Real coefficients since #x^4+3x^2+4 >= 4# for all #x in RR#.

So let's look for quadratic factors. Since there's no term in #x^3#, these must take the form:

#x^4+3x^2+4 = (x^2+ax+b)(x^2-ax+c)#

#=x^4+(b+c-a^2)x^2+a(c-b)x+bc#

Equating coefficients we find:

(i) #b+c-a^2 = 3#

(ii) #a(c-b) = 0#

(iii) #bc = 4#

From (ii), #a = 0# and/or #b = c#.

If #a = 0#, then we are effectively trying to find linear factors of #t^2+3t+4#, where #t = x^2#. This quadratic in #t# has a negative discriminant (#3^2-4*1*4 = 9-16 = -7#), so only Complex zeros.

So try #b=c#:

From (iii), we get #b = c = 2# or #b = c = -2#. If #a in RR# then #a^2 >= 0# and from (i) we get #b+c = 3+a^2 >= 3#. So #b = c = 2#.

Then #a^2 = b+c-3 = 2+2-3 = 1#. So #a = +-1#

Hence #x^4+3x^2+4 = (x^2+x+2)(x^2-x+2)#