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 Algebra Pre-Algebra and Basic Algebra Math Forum

October 16th, 2018, 07:26 PM   #11
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 Originally Posted by SenatorArmstrong The ration room theorem tells me that $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8 ,\pm 12, \pm 16$ are potential solutions. I began by plugging in these values starting with $\pm 1$ when doing so, those values made the equation false. The value of 2 worked. I then went ahead with synthetic division. I went from $z^5 - 3z^4 - 16z + 48$ $\Rightarrow$ $z^4 - 5z^3 + 10z^2 - 20z + 24$ I used the rational root theorem again and found that $-3$ is a root. Proceeding with synthetic division... $z^4 - 5z^3 + 10z^2 - 20z + 24$ $\Rightarrow$ $z^3 - 2z^2 + 4z - 8$ I factored this by method of grouping. $z^2(z-2) + 4(z-2)$ $\Rightarrow$ $(z^2 + 4)(z-2)$ I've been able to conclude that $z=2, -3, \pm 2i$ My original polynomial is a fifth degree polynomial so I was expecting to find one more solution. Where did I go wrong? Kind Regards

The very first root you found is not quite correct. It should be $z=3$ that's a root, and so $(z-3)$ 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
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 Originally Posted by ProofOfALifetime What tipped me off was that he said, "how do you factor" so I sort of made the assumption. I love the rational roots theorem. I especially like how it's used as an alternative way to prove $\sqrt{2}$ is irrational. It's fascinating. (I'm not much of a fan of the proof by contradiction method of $\sqrt{2}$ being irrational.)
Sorry I should have said finding solutions to the polynomial. My apologies for the confusion. So this actually can't be factored? October 16th, 2018, 07:29 PM   #13
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 Originally Posted by SenatorArmstrong $z^5 - 3z^4 - 16z + 48 = 0$ $\Rightarrow$ $z^4(z-3) - 16(z+3)$ The $(z-3)$ and $(z+3)$ looks like a difference of two squares, but don't think I can apply that here. I'm curious how you can factor this quicker since rational root theorem can get time consuming. I appreciate your help.
Check again, your sign is off when you factored from the last two terms. October 16th, 2018, 07:31 PM   #14
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 Originally Posted by ProofOfALifetime Check again, your sign is off when you factored from the last two terms.
I'm going to try this problem again in the morning. I must be making silly mistakes. Thanks for the help though. October 16th, 2018, 07:32 PM   #15
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 Originally Posted by SenatorArmstrong I'm going to try this problem again in the morning. I must be making silly mistakes. Thanks for the help though.
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
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Quote:
 Originally Posted by SenatorArmstrong I've been able to conclude that $z=2, -3, \pm 2i$
z = -2 is also a solution : (-2)^4 = 16 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. Thanks from SenatorArmstrong October 17th, 2018, 09:21 AM   #18
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 Originally Posted by skipjack 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.
Thanks skipjack.

So with $z^4-16 =$ $\Rightarrow$ $(z^2 -4)(z^2+4)$ 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
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 Originally Posted by SenatorArmstrong Thanks skipjack. So with $z^4-16 =$ $\Rightarrow$ $(z^2 -4)(z^2+4)$ 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.
That's what I was trying to say as well! Skipjack explained it better. Tags finding, solutions, stuck 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 bobbo Algebra 8 October 30th, 2013 02:31 PM live504 Algebra 6 April 8th, 2012 09:35 AM grogmachine Algebra 3 August 5th, 2011 01:16 PM blackobisk Algebra 1 April 9th, 2009 02:28 PM live504 Calculus 1 December 31st, 1969 04:00 PM

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