Question 1196478
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Almost certainly the explanation by tutor @ikleyn is the easiest way to show that the given polynomial has no real roots.<br>
But note that Descartes' rule of signs can also be used.<br>
The given polynomial has 0 sign changes, so the number of positive real roots is 0.<br>
The polynomial with x replaced by -x also has 0 sign changes, so the number of negative real roots is also 0.<br>
Along with the fact that x=0 is obviously not a root, that means the number of real roots is 0.<br>