Question: A password is 8 characters long. Each character can contain 26 lower case or 26 uppercase letters or a integer from 0-9. What is the probability that an 8 character password contains exactly 1 OR 2 integers? 

Please explain as the book answer is P = .630 

Assuming that I could use (52^6\*10^2+52^7\*10)/62^8 was incorrect. This is the sum of the product between the possibilities of having one integer in the password and having two integers in the password; 62^8 is the total possibilities given each character can be a lower/uppercase or integer.

Answer: 
Because the equality must hold for all values of $x$, if we let $x=0$ then we get $c=d $, then we let $x=1 $ to get $a+c=b +d$ but our previous findings give us that $a=b $. This generalises to polynomials of higher order, you just compare coefficients.
Assume $a\neq b,c\neq d$. Then, we can write $a=b+D,c=d+E$. Then,
$$bX+DX+d+E=bX+d\\
\implies DX+E=0X+0$$
However, $E$ is constant, and not a function of $X$. Therefore, we need $E=0$, and similarly, we need $D=0$, showing that $a=b,c=d$.