How do you factor completely 5y^8 - 125?
1 Answer
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
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
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
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
y^2-sqrt(5) = y^2-(root(4)(5))^2 = (y-root(4)(5))(y+root(4)(5))
Next we will factor
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
(a^2-sqrt(2)ab+b^2)(a^2+sqrt(2)ab+b^2) = a^4+b^4
Let
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))