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November 3rd, 2012, 12:17 PM  #1 
Math Team Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions  Roots of polynomial
There exists a quadratic polynomial P (x) with integer coefficients such that: (a) Both of its roots are positive integers, (b) The sum of its coefficients is prime, (c) For some integer k, P (k) = ?55. Show that one of its roots is 2, and find the other root. 
November 3rd, 2012, 09:11 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond  Re: Roots of polynomial
P(x) = (x  2)(x  1, P(13) = 55. ax² + bx + c = x²  20x + 36 = 0, a + b + c = 17. (x  m)(x  n) = x²  (m + n)x + mn, 1  m  n + mn = p, n(m  1) = p + (m  1), [p + (m  1)] (mod (m  1)) ? 0 unless m  1 = p, so n = 2, thus 2 must be a root if a = 1. 
November 4th, 2012, 04:04 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,919 Thanks: 2201 
Correct answer for P(x), but the deduction of it isn't quite right. One has p = a(m  1)(n  1), so one of the factors is p and each of the others is 1, which implies one of the roots is 2. Without loss of generality, one can assume that n = 2. It can easily be proved that the other root is not 2, so a = 1. Thus P(x) = (x  m)(x  2), where p = m  1 is prime. Since P(x) = 55 for some integer x, m cannot be odd, and so m > 3. One can now finish off by trying each way in which 55 can be factorized. 
November 4th, 2012, 01:35 PM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond  Re: Roots of polynomial
Second attempt, P(x) = (x  2)(x  1, P(13) = 55. ax² + bx + c = x²  20x + 36 = 0, a + b + c = 17. (ax  m)(x  n) = ax²  anx  mx + mn, a  an  m + mn = p, n(m  a) = p + (m  a), [p + (m  a)] (mod (m  a)) ? 0 unless m  a = p or m  a = 1 (m = 2 if a equals 1, and a cannot be greater than 1 if m  a = 1 since a must divide a + 1 (integer roots), which is only possible if a = 1), so, either n = 2 or m = 2, thus 2 must be a root. Assuming a = 1, (x  2)(x  n) = x²  (n + 2)x + 2n, p = n  1. p ? 5, so p ? {5, 7, 11, 13, 17, . . . } and we find n = 18 for the solution above. 
November 4th, 2012, 05:30 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,919 Thanks: 2201 
To avoid confusion, I suggest using a(x  m)(x  n) instead of (ax  m)(x  n). At some stage, you need to eliminate x²  60x + 116 on the grounds that 1  60 + 116 = 57, which isn't prime. Alternatively, you could explain why primes beyond 17 won't work. 
November 8th, 2012, 10:48 AM  #6  
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond  Re: Roots of polynomial Quote:
uv  nu mv + mn = p n(m  u) = p + v(m  u) m  u divides p + v(m  u) only if m  u = 1 or m  u = p. If m  u = 1, m must be 2 since u must divide u + 1 (integer roots). If m  u = p, then np = p + vp p(n  v) = p so n  v = 1 so v = 1 (integer roots) and n must be two, thus at least one of the roots is 2.  

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