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October 16th, 2018, 08:26 PM   #11
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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.
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October 16th, 2018, 08:28 PM   #12
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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?
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October 16th, 2018, 08:29 PM   #13
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$z^5 - 3z^4 - 16z + 48 = 0$ $\Rightarrow$ $z^4(z-3) - 16(z+3)$

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.
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October 16th, 2018, 08:31 PM   #14
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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.
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October 16th, 2018, 08:32 PM   #15
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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 08:35 PM.
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October 16th, 2018, 08:45 PM   #16
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I've been able to conclude that $z=2, -3, \pm 2i$
z = -2 is also a solution : (-2)^4 = 16
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October 16th, 2018, 10:28 PM   #17
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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
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October 17th, 2018, 10:21 AM   #18
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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.
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October 17th, 2018, 12:47 PM   #19
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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.
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