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

2 Answers
Cesareo R.
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)#

George C.
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)#

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