See discussions, stats, and author profiles for this publication at: https://www.researchgate.net/publication/257821468 Problem 11589: Where are the zeros? Article in The American Mathematical Monthly * July 2011 CITATIONS 0 READS 17 1 author: Some of the authors of this publication are also working on these related projects: Metatheoretical analysis and structural alternatives of the resolving of the philosophical problem of the applicability of mathematics in natural sciences and Wigner's puzzle View project Editing/Proofreading/Commenting/Research Designing in philosophy of science/philosophy of mathematics Collaboration and co-authorship View project Catalin Barboianu University of Bucharest 27 PUBLICATIONS 5 CITATIONS SEE PROFILE All content following this page was uploaded by Catalin Barboianu on 09 May 2014. The user has requested enhancement of the downloaded file. next points are A3 = (5rs : −3sq : −3qr), A4 = (11rs : −5sq : −5qr), and in general An = (anrs : −an−1sq : −an−1qr) = ((an/an−1)rs : −sq : −qr), where {an} is defined recursively by a0 = 1, an = 2an−1 + (−1)n , the Jacobsthal sequence. Take the limit as n → ∞ to find (since an/an−1 → 2) that An → A∗ = (2rs : −sq : −qr), so A∗ is the intersection of p with the line A0 P (r y = sz, same as A0 A1), since its coordinates satisfy both equations. Also solved by R. Chapman (U. K.), M. Goldenberg & M. Kaplan, J.-P. Grivaux (France), A. Habil (Syria), M. E. Kidwell & M. D. Meyerson, L. R. King, O. Kouba (Syria), J. C. Linders (Netherlands), O. P. Lossers (Netherlands), J. Minkus, R. Stong, GCHQ Problem Solving Group (U. K.), University of Louisiana at Lafayette Math Club, and the proposer. Where Are the Zeros? 11589 [2011, 653]. Proposed by Catalin Barboianu, Infarom Publishing, Craiova, Romania. Let P be a polynomial over R given by P(x) = x3 + a2x2 + a1x + a0, with a1 > 0. Show that P has a least one zero between −a0/a1 and −a2. Solution by William J. Cowieson, Fullerton College, Fullerton, CA. Note that P(−a0/a1) = (a20/a31)(a1a2 − a0) and P(−a2) = a0 − a1a0, so that P(−a0/a1) = −(a20/a31)P(−a2). (1) There are three cases. (1) If a0 − a1a2 = 0, then the interval reduces to a single point, and that point is a zero of P . (2) If a0 = 0, then P(x) = x(x2 + a2x + a0) has zeros at 0 and at (−a2 ±√ a22 − 4a1 )/2. If a22 − 4a1 < 0, then 0 is the only real zero of P . Otherwise, (−a2 ± √ a22 − 4a1 )/2 are both strictly between −a0/a1 = 0 and −a2, since a1 > 0. (3) Both a0 − a1a2 = 0 and a0 = 0. In this case, from (1) we see that P(−a0/a1) and P(−a2) are nonzero and of opposite sign when a1 > 0. Hence the Intermediate Value Theorem implies that there is a zero between −a0/a1 and −a2. Also solved by B. K. Agarwal (India), G. Apostolopoulos (Greece), S. J. Baek & D.-H. Kim (Korea), B. D. Beasley, M. W. Botsko, D. Brown & J. Zerger, V. Bucaj, P. Budney, H. Caerols (Chile), E. M. Campbell & D. T. Bailey, M. Can, M. A. Carlton, T. Castro, J. Montero & A. Murcia (Colombia), R. Chapman (U. K.), H. Chen, W. ChengYuan (Singapore), J. Christopher, D. Constales (Belgium), W. J. Cowieson, C. Curtis, P. P. Dályay (Hungary), C. Degenkolb, C. R. Diminnie, K. Farwell, J. Ferdinands, D. Fleischman, V. V. Garcıa (Spain), O. Geupel (Germany), W. R. Green & T. D. Lesaulnier, J.-P. Grivaux (France), M. Hajja (Jordan), E. A. Herman, G. A. Heuer, S. Kaczkowski, B. Kalantari, B. Karaivanov, T. Keller, L. Kennedy, J. C. Kieffer, O. Kouba (Syria), P. T. Krasopoulos (Greece), R. Lampe, K.-W. Lau (China), J. C. Linders (Netherlands), J. H. Lindsey II, O. López (Colombia), O.P. Lossers (Netherlands), J. Loverde, Y.-H. McDowell & F. Mawyer, F. B. Miles, S. Mosiman, K. Muthuvel, M. Omarjee (France), Á. Plaza & K. Sadarangani (Spain), P. Pongsriiam & T. Pongsriiam (U. S. A. & Thailand), V. Ponomarenko, C. R. Pranesachar (India), R. E. Prather, R. Pratt, D. Ritter, A. J. Rosenthal, U. Schneider (Switzerland), C. R. Selvaraj & S. Selvaraj, A. K. Shafie & S. Gholami (Iran), J. Simons (U. K.), N. C. Singer, E. A. Smith, N. Stanciu & T. Zvonaru (Romania), J. H. Steelman, A. Stenger, R. Stong, M. Tetiva (Romania), N. Thornber, V. Tuck & A. Stancu, D. B. Tyler, D. Vacaru (Romania), E. I. Verriest, J. Vinuesa (Spain), T. Viteam (Germany), Z. Vörös (Hungary), M. Vowe (Switzerland), T. Wiandt, H. Widmer (Switzerland), R. Wieler, S. V. Witt, N. Youngberg, J. Zacharias, Z. Zhang, Fejéntaláltuka Szeged Problem Solving Group (Hungary), GCHQ Problem Solving Group (U. K.), University of Louisiana at Lafayette Math Club, Missouri State University Problem Solving Group, NSA Problems Group, and the proposer. February 2013] PROBLEMS AND SOLUTIONS 181 View publication stats