How many choices? (Posted on 2019-02-05)

	Submitted by Brian Smith
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armando has his solution here. Mine is below. Let q(x) be the polynomial resulting from translating p(x) 3 units to the left on the x axis. Then q(x) = p(x+3) and p(3)=-2 implies q(0)=-2. The possible roots of q(x) are then the signed factors of -2: -2, -1, 1, and 2. Reversing the translation then means that 1, 2, 4, and 5 are the possible roots of p(x).

	Subject	Author	Date
	Solution	armando	2019-02-10 03:30:51