The only real root of this equation is irrational.
I graphed the function , and it appears that there is one real root. I wanted to use the rational zero theorem to determine the root but it doesn't agree with my geometric picture. According to Wolfram, if the coefficients of the polynomial are specified to be integers then we can form rational zeros, , from the leading and terminal coefficients.
From this, I got , . But this is incorrect as none of the possible rational zeros check out.
If an integer is a root of the function you mentioned then this root would be a factor of the constant term .However not all the factors of the constant term would be a root of f(x)=0 .I agree with Emakarov the only root of your function is irrational and lies between -2 and -1.
Okay, so because the only real root of this equation is irrational, then the rational zero theorem does not apply? I'm confused because the rational zero theorem doesn't mention that it has these restrictions.
The rational root theorem applied to a polynomial f(x) with integer coefficients says the following.
For every rational number x written as a fraction p / q in lowest terms, if f(x) = 0, then p divides the constant term of f and q divides the leading coefficient of f.
This statement is true for the polynomial from the OP, i.e., . Indeed, take any rational number x. Then f(x) ≠ 0. Therefore, the premise of the implication "if f(x) = 0, then..." is false, which means that the whole implication is true. In other words, the theorem says something only about rational roots of a polynomial. It says nothing about irrational roots, and it does not claim that there have to be rational roots.