Given the expression
-3x^4+27x^2+1200=0
let x^2=a
we can re-write our expression as:
-3a^2+27a+1200=0
-3(a^2-9a-400)=0
a^2-9a-400=0
factorizing the above we have:
a^2+16a-25a-400=0
a(a+16)-25(a+16)=0
(a+16)(a-25)
thus replacing back x^2 we have:
(x^2+16)(x^2-25)
=(x^2+16)(x-5)(x+5)
factorizing (x^2+16) we get
x^2=+/-√-16
x=+/-4i
thus the zeros of the expression are:
x=-5, x=5 , x=-4i, x=4i
