The quartic polynomial $f(x) = x^4 + a x^3 + b x^2 + c x + d$ is such that $ad$ is odd and $bc$ is even. Prove that $f(x)$ does not has all rational roots.
Clearly, $f(x)$ will have either 0 or 2 or 4 rational roots. Let us assume all the four roots of the equation are rational and they be $p_1/q_1, p_2/q_2, p_3/q_3$ and $p_4/q_4$. As the question suggests: a is odd, d is odd, b is even/odd, c is even/odd, but, atleast one of b or c must be even. $(p_1/q_1)*(p_2/q_2)*(p_3/q_3)*(p_4/q_4)=d$ (odd integer) --(1) $(p_1/q_1)+(p_2/q_2)+(p_3/q_3)+(p_4/q_4)=-a$ (odd integer) --(2) if --(1) is true, the denominator of a root must be contained in the numerator of the remaining roots. This means, these numerators cannot have denominators containing a factor of their numerator. So, there will be atleast two denominators whose gcd will be 1. Thus, we cannot find any four rational numbers which add to give an integer because rational numbers only give an integer if their denominators have atleast one common factor. So, the equation cannot have 4 rational roots. Same can be explained for 2 rational roots.
So, I just wanted to ask am I correct in my approach or am I making any mistake. Further, I am not able to understand why there is a distinction of ad and bc as odd and even?? Please help me out..