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

1 Answer
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))