Wednesday 17 February 2016

How to show that a polynomial does not have real roots



How to show generally that a polynomial does not have real roots. Well, for eg lets take the polynomial $x^8-x^7+x^2-x+15$ . Here the power($n=8$) is even so it can have real roots or it might not have real roots.



Something which I thought was to find the minima and show that if the minima of $p(x)$ is greater than $0$ and $a_1$ that is the coefficient of $x^8$ are both greater than $0$ then we cannot have real roots . But in this case the derivative is $8x^7-7x^6+2x-1$ and I cannot find minima for it . So what should I do in this example . Well it is already given this polynomial does not have real roots , but I have to prove it.




Also even if I get that this does not have any real roots then is this a general method for all kinds of polynomials ?



Edit: I know Strum's theorem is one general way to solve such questions but this question is from an undergrad entrance paper and I guess a method under the reach of calculus or something similar will suffice better.


Answer



Clearly there is no negative root as all terms are positive for $x < 0$. The question remains if there are positive roots. Here is a simple way which often works.



Case 1: $0 < x <1$.



$$P(x) = (15-x) + (x^2-x^7) + x^8 > 0$$
as each term is positive.




Case 2: $ x > 1$. Similarly
$$P(x) = (x^8-x^7) + (x^2-x) + 15 > 0$$



as $x=1$ is not a root, we are done.


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