How do you factor #x^4-x^2-12#?

1 Answer
Mar 3, 2016

Answer:

#x^4-x^2-12=(x-2)(x+2)(x^2+3)#

Explanation:

Use the difference of squares identity:

#a^2-b^2=(a-b)(a+b)#

with #a=x# and #b=2# below.

Treat the quartic as a quadratic in #x^2#, noting that #4*3 = 12# and #4-3 = 1#, so:

#x^4-x^2-12#

#=(x^2)^2-(x^2)-12#

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

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

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

which has no simpler factors with Real coefficients, since #x^2+3 >= 3 > 0# for any Real value of #x#. With Complex coefficients we can factor one step more...

#=(x-2)(x+2)(x-3i)(x+3i)#