Show that the polynomial f(x) = …

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Sia ? 6 years, 4 months ago
f(x) = x4 + 4x2 + 6
= (x2)2 + 4x2 + 6
Let x2 =n,
Then, f(x) = n2 + 4n + 6,
Here a=1,b=4,c=6
The discriminant(D) = {tex}\text{b}^2-4\mathrm{ac}=\;(4)^2-4\times1\times6=16-24=-8{/tex}
Since the discriminant is negative so this polynomial has no zeros
Hence, f(x) = x4 + 4x2 + 6 has no zero.
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