October 16th, 2018, 07:26 PM  #11  
Senior Member Joined: Oct 2016 From: Arizona Posts: 193 Thanks: 34 Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems.  Quote:
The very first root you found is not quite correct. It should be $z=3$ that's a root, and so $(z3)$ is a factor. Did you factor by grouping from the beginning? This will make things easier, but as Micrm@ss said, it's sort of a lucky problem.  
October 16th, 2018, 07:28 PM  #12  
Senior Member Joined: Nov 2015 From: United States of America Posts: 195 Thanks: 25 Math Focus: Calculus and Physics  Quote:
 
October 16th, 2018, 07:29 PM  #13  
Senior Member Joined: Oct 2016 From: Arizona Posts: 193 Thanks: 34 Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems.  Quote:
 
October 16th, 2018, 07:31 PM  #14 
Senior Member Joined: Nov 2015 From: United States of America Posts: 195 Thanks: 25 Math Focus: Calculus and Physics  
October 16th, 2018, 07:32 PM  #15 
Senior Member Joined: Oct 2016 From: Arizona Posts: 193 Thanks: 34 Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems.  You're welcome. It's easy to make sign errors on these types of problems.
Last edited by ProofOfALifetime; October 16th, 2018 at 07:35 PM. 
October 16th, 2018, 07:45 PM  #16 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,112 Thanks: 1002  
October 16th, 2018, 09:28 PM  #17 
Global Moderator Joined: Dec 2006 Posts: 20,370 Thanks: 2007 
As the number of terms is even, grouping in pairs should be tried. You made a sign error. The equation can be written as z^4(z  3)  16(z  3) = 0, i.e. (z^4  16)(z  3) = 0, which gives (z^2 + 4)(z + 2)(z  2)(z  3) = 0. Finally, using imaginary numbers, (z + 2i)(z  2i)(z + 2)(z  2)(z  3) = 0. 
October 17th, 2018, 09:21 AM  #18  
Senior Member Joined: Nov 2015 From: United States of America Posts: 195 Thanks: 25 Math Focus: Calculus and Physics  Quote:
So with $z^416 = $ $\Rightarrow$ $(z^2 4)(z^2+4)$ utilizing difference of two squares. I like this approach. A lot quicker than rational root theorem. From here on out, I'll always try grouping first when the terms are even. I'll save rational root theorem as a last resort.  
October 17th, 2018, 11:47 AM  #19  
Senior Member Joined: Oct 2016 From: Arizona Posts: 193 Thanks: 34 Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems.  Quote:
 

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