Maybe I'm misinterpreting but this seems to be trivial. By expanding, $$(x^2-2x+5)(x^2-4x+20)+1=x^4-6x^3+33x^2-60x+101.$$So $3$ does not divide the first coefficient, but does divide the next three, and $3^2=9$ does not divide the constant. By Eisenstein, this is irreducible over the rationals, and therefore over integers as well.

sseraj wrote: By Eisenstein, this is irreducible over the rationals, and therefore over integers as well. In order to use Eisenstein's criterion you have to have $p\mid a_0$ which is not the case here.

Okay, how about this: By expanding, $$(x^2-2x+5)(x^2-4x+20)+1=x^4-6x^3+33x^2-60x+101.$$Since $101$ is prime, the only possible integer roots are $\pm 1, \pm 101$. But substituting these in, we can immediately see that there are no integer roots. So there are no linear factors over $\mathbb{Z}$. So if a factorization is possible then it has to be two quadratics multiplied together like $(x^2+ax+b)(x^2+cx+d)$ for integers $a,b,c,d$. Since $101$ is prime, we can set $a=1$ and $c=101$ so that the polynomial is $$ (x^2+ax+1)(x^2+cx+101)=x^4+(a+c)x^3+(ac+102)x^2+(101a+c)x+101.$$By comparing coefficients, we have $a+c=-6$ and $101a+c=-60$. Subtracting the first from the second gives $100a=-54$ and so $a$ is not an integer. Contradiction.

For P to have a divisor of deg (1) , there must be an integer a , such that P(a) = 0, i.e. (x - a) divides P For negative a , P(a) > 0 , and also for each a >= 6 , P(a) > 0 Checking for 0 <= a <= 5 we also see that P(a) > 0 also. Then we continue with the same approach sseraj did i.e. by saying that the only way that P can be factorised is : P(x) = (ax^2 + bx + c)*(dx^2+ex+f) and we arrive to a condradiction hence P is irreducible

See also problem 2, page 2 here: http://yufeizhao.com/olympiad/intpoly.pdf