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: 
Yes, this can be expressed in the language of linear algebra, as you seem to desire. Namely, over any infinite field $\,F\,$ the polynomials $\,\{X^i\}\,$ are basis of the polynomial function ring $\,K[X],\,$ hence $\ c+ aX = \langle c,a,0,0\ldots\rangle = \langle d,b,0,0\ldots\rangle = d+b X\,$ iff they have the same coefficients, i.e. $\, c=d,\, a=b.\,$ 

Similarly $\,\sin(x),\, \cos(x)\,$ are a basis for the solution space of of $\,y'' + y = 0\,$ (they are linearly independent by the Wronskian criterion) so an analogous statement holds.

**Remark** $\ $ This can fail over *finite* fields, e.g. by little Fermat $\,X^p = X\,$ as *functions* on $\,\Bbb F\_p = \Bbb Z/p.\,$ However, it remains true for the ring of *formal* polynomials $\,R[x]\,$ (vs. polynomial *functions*), since, by definition, two *formal* polynomials are equal iff their corresponding coefficients are equal.
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.