Polynomial with prime exponents

Take 1 to the LHS, then call the expression f(x).

Now, f(x) changes sign once, so there is only one solution.

And, by the similar argument, we get that there are 0 or 2 solutions of the equation. Then, f(0)<0 &f(-1)>0, which implies that there is at least one negative solution, hence there are two of them.

now, the most difficult part,

The total number of the solution are=1+2=?

Descartes Rule of Signs

Intermadiate Value Theorem

$x^{17}+x^{13}+x^{11}+x^{7}+x^{5}+x^{3}+19x^{2}-1=0$

Applying Descartes' Rule of Signs :-

Number of changes in sign of $f(x)$ is 1.

Hence, number of positive roots $=1$

Number of changes in sign of $f(-x)$ is 2

(Because $f(-x)=-x^{17}-x^{13}-x^{11}-x^{7}-x^{5}-x^{3}+19x^{2}-1$

So there are either $0$ or $2$ negative roots.

By, Intermediate value theorem,

There are 2 negative roots

Hence, total no of real roots=3 (0 is not a root)

Hence, answer = $\boxed{3}$

Applying the descartes rule of signs we can get an upper bound of the no. Of positive roots . same concept for (-x) to get upper bound on negative roots , verifying by intermediate value theorem . the no. Of real roots is 3