Number Theory Number Theory Math Forum

 April 17th, 2015, 06:50 AM #1 Newbie   Joined: Apr 2015 From: United States Posts: 8 Thanks: 0 Polynomial divisible by Polynomial Prove that the polynomial $(n^2+2n+1)^3+(n^2+8n+16)^3+(9n^2+42n+49)^3+(9n^2+ 48n+64)^3$is divisible by $2n^2+10n+13$. Last edited by skipjack; April 18th, 2015 at 09:23 PM. April 17th, 2015, 07:29 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2642 Math Focus: Mainly analysis and algebra I think I would show that the two (complex) roots of the divisor are also roots of the dividend. April 17th, 2015, 07:47 AM #3 Newbie   Joined: Apr 2015 From: United States Posts: 8 Thanks: 0 I don't know how to do that. Is there a different way? April 17th, 2015, 08:15 AM #4 Newbie   Joined: Apr 2015 From: United States Posts: 8 Thanks: 0 I've actually got it, thank you for your input! April 18th, 2015, 08:14 PM #5 Senior Member   Joined: Sep 2010 Posts: 221 Thanks: 20 It's rather simple. Each pair of the polynomials to the power 3 has the same divisor $(n^2+2n+1)+(9n^2+48n+64)=(n^2+8n+16)+(9n^2+42n+49 )=10n^2+50n+65=5(2n^2+10n+13)$ Last edited by skipjack; April 18th, 2015 at 09:27 PM. Tags divisible, polynomial Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post sahhwahh Number Theory 9 April 17th, 2015 05:16 PM axerexa Algebra 8 March 26th, 2015 09:46 PM king.oslo Algebra 6 August 27th, 2013 06:11 AM gelatine1 Algebra 4 January 7th, 2013 09:39 PM condemath Algebra 3 September 20th, 2011 08:34 PM

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