How do you factor completely: #2u^3 w^4 -2u^3#?

Question

1352 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-24T17:31:50

#2u^3w^4-2u^3 = 2u^3(w-1)(w+1)(w^2+1)#

First separate out the common factor #2u^3# to get:

#2u^3w^4-2u^3 = 2u^3(w^4-1)#

Then use the difference of squares identity (twice):

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

#2u^3(w^4-1)#

#=2u^3((w^2)^2-1^2)#

#=2u^3(w^2-1)(w^2+1)#

#=2u^3(w^2-1^2)(w^2+1)#

#=2u^3(w-1)(w+1)(w^2+1)#

#w^2+1# has no linear factors with real coefficients since #w^2+1 >= 1 > 0# for all #w in RR#