If $ad$ and $bc$ are odd and even, respectively, then prove that $ax^3+bx^2+cx+d$ has an irrational root. $\displaystyle ax^3+bx^2+cx+d$ is a polynomial with integer coefficients. It is given that $ad,\,bc$ are odd and even respectively. Then prove that not all roots of the polynomial are rational.
It is easy to see that none of the roots are integer. But how to tackle the rational case? Any help is appreciated.
 A: Hint $ $ If not, then it has $\,3\,$ rational roots of form $\,r/s\,$ for $\rm\color{#c00}{odd}$ $\,r,s,\,$ by the Rational Root Test.  These rational roots persist modulo $2$ as roots $\,r/s\equiv \color{#c00}1/\color{#c00}1\equiv 1,\,$ therefore mod $\,2\,$ the polynomial $\equiv (x-1)^3\equiv x^3+x^2+x+1,\,$ thus $\,b\equiv 1\equiv c$ are both odd, contra $\,bc\,$ even.
A: Let us assume that  $\alpha,\beta,\gamma$  are the roots of the polynomial $p(x)$  and are also rational for which we can say $\alpha=\frac{a_2}{a_1},\beta=\frac{b_2}{b_1} and   \gamma=\frac{c_2}{c_1}$ where $a_1,b_1,c_1,a_2,b_2,c_2$ are integers.
Therefore we can write $p(x)$ as $$kp(x)=(a_1x+a_2)(b_1x+b_2)(c_1x+c_2)$$ $$\implies a=a_1b_1c_1$$ $$b=a_1b_1c_2+a_1b_2c_1+a_2b_1c_1$$ $$c=a_1b_2c_2+a_2b_1c_2+a_2b_2c_1$$ $$d=a_2b_2c_2 $$ (neglecting $k$)
. 
As $ad$ is odd, therefore $a_1,b_1,c_1,a_2,b_2,c_2$ are all odd individually.
This implies that $b$ and $c$ must also be odd individually, which is a contradiction. 
Hence not all the roots can be represented in the above way which signifies that all the roots cannot be rational.
