Question 58903
f(x) = x^4 + 5x^2 + 4
Let x^4 + 5x^2 + 4 = 0..............(1)
Let x^2 = y, then x^ 4 = y^2
So eqn (1) becomes y^2 + 5y + 4 = 0
           ==>     y^2 + y + 4y + 4 = 0 [Splitting the middle term]
           ==>     y(y+1) + 4(y+1) = 0
           ==>     (y+1) (y+4) = 0
           ==>     y+1 = 0   or  y+4 = 0
           ==>     y = -1 [adding -1 to both the sides] or y = -4 [Adding -4]
           ==>     x^2 = -1  or x^2 = -4 [because y = x^2]
           ==>     x = sqrt(-1)  or x = sqrt(-4)  [taking sqrt]
We know that sqrt(-1) = i
Therefore sqrt(-1) = i 0r -i and sqrt(-4) = 2i or -2i

Thus the zeroes of the given function are i, -i, 2i and -2i

f(x) = x^4 + 5x^2 + 4
     = (x+i)(x-i)(x+2i)(x-2i)

Regards,
Uma