For a polynomial with integer coefficients, is it true that if constant term is prime then it cannot be the root of the polynomial. 
For a polynomial with integer coefficients, is it true that if constant term is prime then it cannot be the root of the polynomial.



Let $p$ be a polynomial with constant term $a_0$ and if $a_0$ is prime then $p(a_0) \ne  0$ 


I just thought of this while working on some other problem.
Is this true ?
My attempt :- 
$$|a_0/a_n| = |r_1 ... r_n|$$
If, $r_1 = a_0$
$$1 = |a_n||r_2 ... r_n|$$
Also $$|a_1/a_n| = |r_2...r_n + r_1r_3...r_n + ... + r_1...r_{n-1}|$$
Or $$|a_1| =  |r_1|\left| \dfrac{1}{r_1} + \dfrac{1}{r_2} + \cdots + \dfrac1{r_n}\right|$$
I am having difficulty in proving that $\left| \dfrac{1}{r_1} + \dfrac{1}{r_2} + \cdots + \dfrac1{r_n}\right|$ not a integer if $|r_1 ... r_n|^{-1}$ is a integer.
Any hints ? 

Ok this is false but can somebody prove/disprove this if $p(1) \ne p(-1) \ne  0$ ?

Consider the polynomial $$f(x)=x^2 - 6 x + 5$$
The constant term here is $5$, which is prime.
However, \begin{align}f(5)&=5^2-6\times 5+5\\
&=25-30+5\\
&=0\end{align}
And thus $5$ is a root of the equation.
Therefore, I have disproved your hypothesis through contradiction

For higher degree polynomials, anything of the form $$f(x)=(x\pm p)(x+1)^a(x-1)^b$$ for $p$ prime, and $a,b$ integers will form a contradiction.
Examples include $$x^4 + 8 x^3 + 6 x^2 - 8 x - 7 = (x+7)(x+1)^2(x-1)$$ which has $p=-7$, $a=2$, $b=1$
and $$x^6 - 14 x^5 + 11 x^4 + 28 x^3 - 25 x^2 - 14 x + 13 = (x-13)(x+1)^2(x-1)^3$$ which has $p=13$, $a=2$, $b=3$

To add the constraint that $f(1)\neq f(-1)\neq 0$, then we can choose $$x^3 - 11 x^2 + x - 11=(x-11)(x^2+1)$$ 
Here we have $f(1)=-20$, $f(-1)=-24$ but $f(11)=0$
As long as the second polynomial is an irreducible one, with a constant of $1$, then this will form a contradiction
