Originally Posted by
Soroban Hello, ZetaX!
I would assume you're expected to use the Rational Roots Theorem.
Given a polynomial $\displaystyle p(x)$, if $\displaystyle p(a) = 0$, then $\displaystyle (x-a)$ is a factor of $\displaystyle p(x).$
All the polynomials have a leading coefficient of 1.
. . So the only possible rational roots are factors of the constant term.
For (1), the only choices are: .$\displaystyle \pm1\quad\hdots $ Try them in $\displaystyle f(x).$
For (2), the only choices are: .$\displaystyle \pm1, \pm2, \pm3, \pm4, \pm6, \pm12$
For (3), the only choices are: .$\displaystyle \pm1, \pm2, \pm3, \pm4, \pm6, \pm8, \pm12, \pm24$
For (4), the only choices are: .$\displaystyle \pm1$