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
