Polynomial factorization

Question

How can I completely factorize y⁸ + 64 ?

Answer 1

y^8 + 64

= (y^4 -4y^2 +8)(y^4+4y+8)

=y^4(y^4 +4y^2+8) -4y^2(y^4 +4y^2 +8) + 8(y^4+4y^2+8)

= y^8 +4y^6 +8y^4 -4y^6 -16y^4 -32y^2 + 8y^4 +32y^2 + 64

all terms cancel except the y^8 + 64

2 y^6 terms cancel each other +4y^6 -4y^6

3 y^4 terms cancel +8y^4-16y^4+8y^4

2 y^2 terms cancel +32y^2-32y^2

because y^8+64 doesn't resemble standard alegebraic identities

many knowledgeable talented mathematicians are tempted to deny it can be factored at all

there are other factors, but they're likely non integer factors such as using sum of cubes formula

(a^3+b^3)= (a+b)(a^2 -ab+ b^2), with b=4, a= x^(8/3) = (x^(1/3)^8= x^2(x^(1/3)^2

=(x^2(x^(1/3)^2+ 4)(x^2(x^(1/3)^4 +4)(x^5x^(2/3)^2 +16) which generally is non-integer, unless x=1 or similar values

with x=1 =a, b=4

=(a+b)(a^2-ab+b^2)= (1+4)(1-4+16) =5(13)=65

=a^3+ b^3 = (1+4^3) =(1+64) =65

so you could use sum of cubes formula with restrictions on x, and factor it into integer factors

factoring is often trial and error, and a trying trial sometimes

one way around that is

use the factoring calculators on line such as Mathpapa or Symbolab

which are both free, some cost money

one though mistakenly says it's not factorable

the other correctly gives the above 1st solution the product of 2 integer trinomials

Answer 2

y^8 + 64 does factor as

(y^4 + 4y^2 + 16) * (y^4 -4y^2 + 16)

FOIL is a very bad method for multiplying trinomials. The box method is much better.

The above product can be factored more if complex numbers are allowed.

TIMES Y^4 -4Y^2 8

Y^4 Y^8 -4Y^6 8Y^4

4Y^2 4Y^6 -16Y^4 32Y^2

8 8Y^4 32Y^2 64

The intersecting rows and columns represent the product of the monomials in the top row and leftmost column.

Note that most of the diagonal elements cancel out, leaving only y^8 + 64. Anyone still believe that this product is incorrect?