Trial and error, We know from Descartes rule of signs that there is exactly one positive root, and 0 or 2 negative roots. Also we know that for large positive x f(x) is positive, and "large" negative x f(x) is negative from the sign of the x^3 term.

So evaluate f(0) it should be -ve, take positive steps untill f changes sign (in this case between 1 and 2). Then as the sum of the roots is equal to -the coefficient of x^2, we conclude that the negative roots if they exist, are >=-5.

There are almost certainly other methods of establishing a suitable range.

RonL