How do you factor #5y^6-5y^2#?

Question

1209 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-28T15:57:01

First separate the common factor #5y^2# then factor the remaining well known quartic...

#5y^6-5y^2#

#=5y^2(y^4-1)#

#=5y^2(y-1)(y+1)(y^2+1)#

The difference of squares identity is:

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

We use this twice below...

First separate the common factor #5y^2# then factor the remaining quartic:

#5y^6-5y^2#

#=5y^2(y^4-1)#

The quartic factor is a difference of squares...

#=5y^2((y^2)^2-1^2)#

#=5y^2(y^2-1)(y^2+1)#

The first quadratic factor is a difference of squares...

#=5y^2(y^2-1^2)(y^2+1)#

#=5y^2(y-1)(y+1)(y^2+1)#

The remaining quadratic factor has no linear factors with real coefficients since #y^2+1 >= 1 > 0# for all #y in RR#