I was thinking rational root theorem. But when I look at the solution it uses three back to back iterations of synthetic division, and I got slightly confused. Any tips?

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I was thinking rational root theorem. But when I look at the solution it uses three back to back iterations of synthetic division, and I got slightly confused. Any tips?

Hi SenatorArmstrong, you should focus on what you can factor out of the first two terms. Then, can you factor something out of the last two terms? Try it out! (Hint: you'll need to factor as much as possible, so after your first round of factoring, check to see if any of the factors can be factored further)

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You all are making it more complicated than need be. A simple factor by grouping works, and then factoring again.OK, you mentioned the rational root theorem. Very good start!

So if you apply the rational root theorem to this polynomial, what do they say the rational roots are?

I was just kiddingYou all are making it more complicated than need be. A simple factor by grouping works, and then factoring again.

You are definitely right. I missed that. It is indeed a lot simpler. But these are tricks that don't always work and where you need to be quite lucky. A rational root theorem is much more generally applicable and gives you all the rational root. Of course, it still isn't general enough for the general quintic.You all are making it more complicated than need be. A simple factor by grouping works, and then factoring again.

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.)You are definitely right. I missed that. It is indeed a lot simpler. But these are tricks that don't always work and where you need to be quite lucky. A rational root theorem is much more generally applicable and gives you all the rational root. Of course, it still isn't general enough for the general quintic.

The rational room theorem tells me that $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8 ,\pm 12, \pm 16$ are potential solutions.OK, you mentioned the rational root theorem. Very good start!

So if you apply the rational root theorem to this polynomial, what do they say the rational roots are?

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?

Thanks a lot for the help.

Kind Regards

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$z^5 - 3z^4 - 16z + 48 = 0$ $\Rightarrow$ $z^4(z-3) - 16(z+3)$Hi SenatorArmstrong, you should focus on what you can factor out of the first two terms. Then, can you factor something out of the last two terms? Try it out! (Hint: you'll need to factor as much as possible, so after your first round of factoring, check to see if any of the factors can be factored further)

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.