Question

How do you factor `x^4+2x^3y-3x^2y^2-4xy^3-y^4`?

Answer 1

`(x-(1+sqrt(5))y/2)(x-(1-sqrt(5))y/2)`
`(x+(3+sqrt(5))y/2)(x-(sqrt(5)-3)y/2)=0`

Explanation:

`"Solve the characteristic quartic equation without the y's first :"`
`x^4 + 2 x^3 - 3 x^2 - 4x - 1 = 0`
`=> (x^2-x-1)(x^2+3x+1) = 0" (*)"`
`"1) "x^2+3x+1=0 => x = (-3 pm sqrt(5))/2`
`"2) "x^2-x-1 = 0 => x = (1 pm sqrt(5))/2`

`"If we apply this on the given polynomial we get"`
`(x^2 - x y - y^2)(x^2 + 3 xy + y^2) = 0`
`=> (x-(1+sqrt(5))y/2)(x-(1-sqrt(5))y/2)`
`(x+(3+sqrt(5))y/2)(x-(sqrt(5)-3)y/2)=0`

`"(*) With the substitution "x = y-1/2" we get : "`
`y^4 - (9/2) y^2 + 1/16 = 0`
`"Now put "z = y^2" and multiply with 16 : "`
`16 z^2 - 72 z + 1 = 0`
`"disc : "72^2 - 4*16 = 5120 = 32^2 * 5`
`=> z = (72 pm 32 sqrt(5))/32 = 9/4 pm sqrt(5)`
`=> y = pm sqrt(9/4 pm sqrt(5))`
`=> y = pm sqrt(9 pm 4 sqrt(5))/2`
`=> y = pm sqrt((2 pm sqrt(5))^2)/2`
`=> y = pm(1 pm sqrt(5)/2)`
`=> x = (1 pm sqrt(5))/2 " or "(-3 pm sqrt(5))/2`