4. How many polynomials $p(x)$ of second degree, with integer coefficients and 2 integer roots, are there such that $p(8)=1$? (Note: remember that integers can be positive, negative, or zero)
(A) 1
(B) 2
(C) 3
(D) a finite number greater than 3
(E) infinitely many.
4. The answer is (C). Let $m, n$ be the two roots (possibly coincident) of the polynomial, then $p(x)=$ $a(x-m)(x-n)$, where $a$ is the coefficient of $x^{2}$, and is therefore also an integer. Thus, $1=p(8)=a(8-$ $m)(8-n)$. But 1 can be obtained as the product of three integers only if they are all 1, or two are -1 and the other is 1. Therefore, the possibilities are $a=1, m=n=7$ or $a=1, m=n=9$ or $a=-1, m=7, n=9$ (NB: swapping $m$ with $n$ does not change the polynomial). Therefore, there is one polynomial with two distinct roots and two other polynomials with coincident roots.