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2001 AMC 12 Problem 23

Problem 23 of 25HarderAlgebra

A polynomial of degree four with leading coefficient 11 and integer coefficients has two real zeros, both of which are integers. Which of the following can also be a zero of the polynomial?

Answer choices

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Solution

Writing P(x)P(x) =(x−r)(x−s)(x2+αx+β)= (x - r)(x - s)(x^2 + \alpha x + \beta) with integer roots r,s,r, s, matching coefficients forces α\alpha and β\beta to be integers. The other two zeros are −α2±i4β−α22. -\dfrac{\alpha}{2} \pm \dfrac{i\sqrt{4\beta - \alpha^2}}{2}. A real part of 12\dfrac{1}{2} requires α=−1,\alpha = -1, making the imaginary part 4β−12.\dfrac{\sqrt{4\beta - 1}}{2}. Choice A needs 4β−1=11,\sqrt{4\beta - 1} = \sqrt{11}, i.e. β=3,\beta = 3, an integer, so it works. The other choices force a non-integer β\beta (for example choice E needs β=3.5,\beta = 3.5, and choice D needs α=−2\alpha = -2 with β=54\beta = \tfrac{5}{4}). Thus, the correct answer is A.
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Tagged: polynomial · complex number · quadratic

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