how do you factor x^15+y^15 over integers?

Question

its for my math homework and im stuck, please help

Answer 1

Greetings,

Consider f(x) = x^15 + y^15 as a function of x

Then f(-y) = -y^15 + y^15 = 0

so (x + y) is a factor and we can find the other factor by division.

-y|1..0..0..0..0..0..0..0..0..0..0..0..0..0..0..y^15
..|...-y..y^2..-y^3..y^4..-y^5..y^6..-y^7..y^8..-y^9..y^10
...1..-y..y^2..-y^3..y^4..-y^5..y^6..-y^7..y^8..-y^9..y^10

The other factor is (x^14 - x^13y + x^12y^2 - x^11y^3 ...+y^14)

with the the general term x^(14-r)(-y)^r where r is a natural number 0<= r <= 14

Regards

Answer 2

You can write; x^15+y^15 = (x^5)³ + (y^5)³

= (x^5 + y^5)(x^10 - x^5y^5 + y^10), by applying formula for
the factors of the sum of two cubes.

Now by the Remainder Theorem, or otherwise, (x + y) is a factor of:

(x^5 + y^5) = (x + y)(x^4 - x³y + x²y² - xy³ + y^4).

Hence, finally, complete factors over integers are:

x^15+y^15

= (x + y)(x^4 - x³y + x²y² - xy³ + y^4)(x^10 - x^5y^5 + y^10).

Answer 3

If you want the complete factorization and you want to do it
without computers , a quick start is to factor out the obvious prime factors which are,

(x+y)(x^2-xy+y^2)(x^4-x^3y+x^2y^2-xy^3+y^4). These come

from the fact that x^3+y^3 and x^5+y^5 divide x^15+y^15.
as well as x+y. After some work, or a TI-89, the fourth factor
is,

(x^8 + x^7y - x^5y^3 - x^4y^4 - x^3y^5 +x^7y +y^8).

Sorry, factor number three did not come through, it is,

(x^4-x^3y+x^2y^2-xy^3+y^4)

Answer 4

x^15+y^15
= (x^5)^3 + (y^5)^3
= (x^5 + y^5)(x^10 - x^5 y^5 + y^10)
= (x + y)(x^4 - x^3 y + x^2 y^2 - x y^3 + y^4)(x^10 - x^5 y^5 + y^10)

Answer 5

= ((x^5)^3 + y^5)^3)
Now use the sum of tw0 cubes formula to factor this.