How do you factor x^4+2x^3y−3x^2y^2−4xy^3−y^4 without quadratic equation?

2 Answers
Mar 4, 2018

-(y+1/2(1+sqrt5)x)(y+1/2(1-sqrt5)x)(y+1/2(3+sqrt5)x)(y+1/2(3-sqrt5)x)

Explanation:

Given

f(x,y)=x^4 + 2 x^3 y -3 x^2 y^2 - 4 x y^3 - y^4

now making the substitution y = lambda x

f(x,lambda x) = (1-lambda-lambda^2)(1+3lambda+lambda^2) x^4

but

1-lambda-lambda^2 = -(lambda+1/2(1+sqrt5))(lambda+1/2(1-sqrt5))

1+3lambda+lambda^2 = (lambda+1/2(3+sqrt5))(lambda+1/2(3-sqrt5)) and then

f(x,y) = -(y+1/2(1+sqrt5)x)(y+1/2(1-sqrt5)x)(y+1/2(3+sqrt5)x)(y+1/2(3-sqrt5)x)

Jul 28, 2018

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

=-(y+(1/2-sqrt(5)/2)x)(y+(1/2+sqrt(5)/2)x)(y+(3/2-sqrt(5)/2)x)(y+(3/2+sqrt(5)/2)x)

Explanation:

Given:

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

Since this is a homogeneous polynomial, factoring it is similar to factoring a corresponding quartic polynomial in one variable.

To keep it reasonably tidy, reverse the order of the coefficients and their signs to get:

z^4+4z^3+3z^2-2z-1

= (z^4+4z^3+6z^2+4z+1)-3(z^2+2z+1)+1

= (z+1)^4-3(z+1)^2+1

= t^4-3t^2+1" " where t = z+1

= t^4-2t^2+1-t^2

= (t^2-1)^2-t^2

= ((t^2-1)-t)((t^2-1)+t)

= (t^2-t-1)(t^2+t-1)

= ((z+1)^2-(z+1)-1)((z+1)^2+(z+1)-1)

= (z^2+2z+1-z-1-1)(z^2+2z+1+z+1-1)

= (z^2+z-1)(z^2+3z+1)

= ((z+1/2)^2-5/4)((z+3/2)^2-5/4)

= ((z+1/2)-sqrt(5)/2)((z+1/2)+sqrt(5)/2)((z+3/2)-sqrt(5)/2)((z+3/2)+sqrt(5)/2)

= (z+1/2-sqrt(5)/2)(z+1/2+sqrt(5)/2)(z+3/2-sqrt(5)/2)(z+3/2+sqrt(5)/2)

Hence:

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

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

=-(y+(1/2-sqrt(5)/2)x)(y+(1/2+sqrt(5)/2)x)(y+(3/2-sqrt(5)/2)x)(y+(3/2+sqrt(5)/2)x)