Show that the polynomial $f (x) = x^{4} + 4 x^{2} + 6$ has no zero.

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Solution

Let $x^{2} = t$ then the given equation reduces to
$t^{2} + 4 t + 6 = 0 D = b^{2} - 4 a c D = 4^{2} - 4 (6) (1) D = 16 - 24 = - 8 D < 0$
Discriminant is less than zero,so the given equation has imaginary roots.

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