All about roots

Let us assume that there are three rational roots of $f(x)$ , namely $\dfrac{p_1}{q_1},\dfrac{p_2}{q_2}$ and $\dfrac{p_3}{q_3}$ where we have WLOG $q_1,q_2,q_3\gt 0$ and $\gcd(p_i,q_i)=1~\forall~1\leq i\leq 3$

Now, by Vieta's formulas, we have,

\displaystyle\frac{p_1}{q_1}+\frac{p_2}{q_2}+\frac{p_3}{q_3}=-\frac ba\implies a(p_1q_2q_3+p_2q_3q_1+p_3q_1q_2)=-bq_1q_2q_3\tag 1

\displaystyle\frac{p_1p_2}{q_1q_2}+\frac{p_2p_3}{q_2q_3}+\frac{p_2p_2}{q_1q_2}=\frac ca\implies a(p_1p_2q_3+p_2p_3q_1+p_3p_1q_2)=cq_1q_2q_3\tag 2

Now, since $ad$ is odd, both $a,d$ must be odd and since $bc$ is even, either $b$ or $c$ is even.

By the rational root theorem and since both $a,d$ are odd, we know that all the $p_i$ 's and $q_i$ 's are odd and hence any product involving the $p_i$ 's and the $q_i$ 's will be odd.

If $b$ is even, we consider $(1)$ for which the LHS is odd $\times$ odd $=$ odd and the RHS is even $\times$ odd $=$ even which is a contradiction since LHS and RHS cannot have different parity.

If $c$ is even, we consider $(2)$ and use a similar argument to show that the parity of the LHS and the RHS is different which is a contradiction.

Hence, in both cases, we arrive at a contradiction and hence our assumption that $f(x)$ has all rational roots was wrong.

Therefore, $f(x)$ cannot have all rational roots.