Polynomials with integer coefficients Through definitions, theorems and my professor the following is true:


*

*The product of any two odd integers is odd. 

*The sum and difference of any two odd integers are even.

*The sum, product and difference of any two even integers are even. 


When expression of the form $(x-r)(x-s)$ are multiplied out a quadratic polynomial is obtained. 
For example $(x-2)(x+7)$ = $x^2 + 5x -14$
But I cannot explain how $x^2 – 1253x + 255$ can’t be written as a product of two polynomials with integer coefficients.  
 A: If you could write $x^2-1253x+255=(x-r)(x-s)$, you would have $rs=255, r+s=1253$  Given that $255=3\cdot 5 \cdot 17$ there are not many choices for $r,s$ and none of them add up to $1253$.
A: The key is to use parity, as hinted. $\rm\:mod\ 2\!:\ f(x) \equiv x^2 + x + 1\:$  has no roots, since $\rm\ f(0)\equiv 1\equiv f(1).\:$ Hence there are no integer roots (else reducing mod $\,2\,$ yields a root mod $\,2$). More generally:
Parity Root Test $\ $ A polynomial $\rm\:f(x)\:$ with integer coefficients 
has no integer roots if its constant coefficient and coefficient sum are both odd. 
Proof $\ $ The test  verifies that $\rm\  f(0) \equiv  1\equiv f(1)\ \ (mod\ 2)\:,\ $ i.e. 
that $\rm\:f(x)\:$ has no roots modulo $2$, hence no integer roots. $\ $ QED
This test extends to many other rings which have a "sense of parity", i.e. an image $\cong \Bbb Z/2,\:$ for example, various algebraic number rings such as the Gaussian integers.
A: Suppose there are integers $r,s$ such that $rs=255, \ r+s=1253$.
From $rs=255$ we get that both $r$ and $s$ are odd and from $r+s=1253$ a contradiction.
A: There are several ways to solve this problem...
1/ One way is to use the Rational Roots Theorem, which said if I can find a ration root for the given polynomial, that means the polynomial is reducible in the set of rationals (thus in the set of integers).  But a simpler way to check is that if there is such a root, say r, exists, then r must divides the "constant" coefficient, which is 255.  Now we can list all the possibilities for the divisors, plug them in and try them all out...
2/ Or, we can use the idea in Eric's answer
3/ The long way (and may be impossible), is to use the Eisenstein' criterion.  The idea is to come up with a prime p such that it divides all coefficients (except the one of the highest power term), it doesn't divides the highest-power coefficient, and p^2 doesn't divide the last "constant" coefficient.  If such prime exists, then we say the polynomial is not reducible (i.e. can't factor) over the field of rationals (thus, over the set of integers)
I said this may be impossible because sometimes there is no such prime exist.
A: $$x^2 – 1253x + 255=(x-x_1)(x-x_2)$$ where $$x_{1,2}=\frac{1253\pm\sqrt{1253^2-4\cdot255}}{2}=\frac{1253\pm 1252.59}{2}\notin\mathbb Z$$
A: Well, if $x^2 - 1253x + 255 = (x - a) (x - b)$ for two integers $a$ and $b$, then $ab = 255$ and $a + b = 1253$. The number $255 = 3*5*17$, so there are only a few possibilities:
$a = 1, b = 255$
$a = 3, b = 85$
$a = 15, b = 17$
$a = 85, b = 3$
$a = 255, b = 1$
And clearly none of those satisfy $a + b = 1253$. More generally, there's a theorem called the rational root theorem which tells you when a given polynomial with integer coefficients has a rational/integer root:
http://en.wikipedia.org/wiki/Rational_root_theorem
