Question

Find all the zeros of the equation. -3x^4 + 27x^2 + 1200 = 0

Answer 1

-3x^4 + 27x^2 + 1200 = 0

factor out -3

-3(x^4 -9x^2 -400)=0

divide by -3

x^4 -9x^2 -400=0

(x^2-25) (x^+16) = 0

x^2-25=0     x^2+16=0

(x-5)(x+5)=0      x^2=-16

x=5   x=-5           x = 4i   x=-4i

Answer:  x=5   x=-5           x = 4i   x=-4i

Answer 2

Answer:

The answer is 5,-5,4i,-4i.

Step-by-step explanation:

When the equation is divided by -3 it becomes:

$x^4-9*x^2-400=0$

If we factorize the equation, the equation will become as:

$(x+5)*(x-5)*(x^2+16)=0$

That means the equation has 2 real zeros.

$x=5\\x=-5$

Also it has 2 complex zeros also. Those are:

$x=-4i\\x=4i$