The polynomial $x^{4} + 4 x^{2} + 5$ has no real zero. If true enter 1, else 0.

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Solution

A polynomial has no zeros when its discriminant D is less than zero.
Here, $x^{4} + 4 x^{2} + 5$ .
Then, $D = b^{2} - 4 a c$
$= 4^{2} - 4 \times 1 \times 5$
$= 16 - 20$
$= - 4$
$< 0$
Thus, the polynomial $x^{4} + 4 x^{2} + 5$ has no zeros.
Hence, the statement is true.

Therefore, the answer is $1$ .

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