Not ( x y ) 2 (x-y)^2

Algebra Level 4

Consider the polynomial x 4 + y 4 + z 4 2 x 2 y 2 2 y 2 z 2 2 z 2 x 2 . x^4 + y^4 + z^4 - 2x^2y^2 - 2y^2z^2 - 2 z^2 x^2.

Can it be factored over the integers (polynomials have integer coefficients)?

Yes it can be completely factored into 4 linear factors. Yes, it can be completely factored into a linear and a cubic factor. Yes, it can be completely factored into 2 quadratic factors. No, it is irreducible over the integers.

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1 solution

x 4 + y 4 + z 4 2 x 2 y 2 2 y 2 z 2 2 z 2 x 2 x^4 + y^4 + z^4 - 2x^2y^2 - 2y^2z^2 - 2z^2x^2

( x 2 + y 2 ) 2 4 x 2 y 2 + z 4 2 z 2 ( x 2 + y 2 ) (x^2+y^2)^2 - 4x^2y^2 + z^4 - 2z^2(x^2 + y^2)

( x 2 + y 2 z 2 ) 2 4 x 2 y 2 (x^2 + y^2 - z^2)^2 - 4x^2y^2

( x 2 + y 2 z 2 2 x y ) ( x 2 + y 2 z 2 + 2 x y ) (x^2 + y^2 - z^2 - 2xy)(x^2 + y^2 - z^2 + 2xy)

( ( x y ) 2 z 2 ) ( ( x + y ) 2 z 2 ) ((x-y)^2 - z^2)((x+y)^2 -z^2)

( x y z ) ( x y + z ) ( x + y z ) ( x + y + z ) (x - y - z)(x - y + z)(x + y -z)(x + y + z)

( y + z x ) ( x + z y ) ( x + y z ) ( x + y + z ) -(y + z - x)(x + z - y)(x + y -z)(x + y + z)

That is an amazing solution!!!!

Tanya Gupta - 7 years, 3 months ago

good solution

VISHESH KUSHWAH - 7 years, 3 months ago

yes a good solution

Rohan Kumar - 7 years, 3 months ago

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