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Algebra Difficulty 3.9 AMC 10/12 Find the answer Italy

Problem:

How many second-degree polynomials $p(x)$ , with integer coefficients and with 2 integer roots, are there such that $p(8)=1$ ? (Note: recall that integers can be positive, negative, or zero)

Pick one

Solution

Solution:

The answer is (C). Letting $m, n$ be the two roots (possibly coincident) of the polynomial, we have $p(x) = a(x-m)(x-n)$ , where $a$ is the coefficient of $x^{2}$ , hence also an integer. Therefore $1 = p(8) = a(8-m)(8-n)$ . But $1$ can be obtained as a product of three integers only if all three are $1$ , or if two are $-1$ and the other is $1$ . So we have the possibilities $a=1, m=n=7$ or $a=1, m=n=9$ or $a=-1, m=7, n=9$ (NB: swapping $m$ with $n$ the polynomial does not change). Hence there is one polynomial with two distinct roots and two other polynomials with two coincident roots.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.