How do you factor completely #5y^8 - 125#?

1 Answer
George C.
Apr 24, 2016

#5y^8-125#

#=5(y^4-5)(y^4+5)#

#=5(y-root(4)(5))(y+root(4)(5))(y^2+sqrt(5))(y^2-root(4)(20)y+sqrt(5))(y^2+root(4)(20)y+sqrt(5))#

Explanation:

First note that both terms are divisible by #5#, so separate that out as a factor:

#5y^8-125 = 5(y^8-25)#

Next we will use the difference of squares identity, which may be written:

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

with #a=y^4# and #b=5#, as follows:

#y^8-25 = (y^4)^2-5^2 = (y^4-5)(y^4+5)#

Next use the difference of squares identity again, this time with #a=y^2# and #b=sqrt(5)# as follows:

#y^4-5 = (y^2)^2-(sqrt(5))^2 = (y^2-sqrt(5))(y^2+sqrt(5))#

Next use the difference of squares identity with #a=y# and #b=root(4)(5)# as follows:

#y^2-sqrt(5) = y^2-(root(4)(5))^2 = (y-root(4)(5))(y+root(4)(5))#

Next we will factor #(y^4+5)#

First notice that:

#(a^2-kab+b^2)(a^2+kab+b^2) = a^4+(2-k^2)a^2b^2+b^4#

In particular, if #k = sqrt(2)# then

#(a^2-sqrt(2)ab+b^2)(a^2+sqrt(2)ab+b^2) = a^4+b^4#

Let #a=y#, #b=root(4)(5)# and #k=sqrt(2)#.

Then:

#y^4+5 = y^4 + (root(4)(5))^4#

#=(y^2-sqrt(2)root(4)(5)y+sqrt(5))(y^2+sqrt(2)root(4)(5)y+sqrt(5))#

#=(y^2-root(4)(20)y+sqrt(5))(y^2+root(4)(20)y+sqrt(5))#

Putting it all together:

#5y^8-125 = 5(y-root(4)(5))(y+root(4)(5))(y^2+sqrt(5))(y^2-root(4)(20)y+sqrt(5))(y^2+root(4)(20)y+sqrt(5))#

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