Factor the polynomial, applying the different types of factoring. How would you go about this?
p^3 - 3p^2 - 27 = 0

Question

Factor the polynomial, applying the different types of factoring.  How would you go about this?
p^3 - 3p^2 - 27 = 0

nubeer · Answer

you have to find the first factor by hit and trial method.. do u know of it?

anonymous · Answer

No.  I don't see any common factors that can come out.

anonymous · Answer

27 = 3 * 3 * 3. I would try to make something out of that. But I can't see it right away.

anonymous · Answer

This can't be factored in an obvious way. p = 3, or 9 are both not roots.

anonymous · Answer

See that's what I was thinking.  It can't be factored.

anonymous · Answer

It can be factored actually. If you take the first derivative, you'll find that it has two roots. And then you can use those to factor it.

anonymous · Answer

What do you mean?  What's the first derivative?  It's been awhile since I've worked with these.

anonymous · Answer

Oh, maybe I was saying things that are wrong actually... I haven't done these things for a while either.

anonymous · Answer

Am I on the right track with this.
p^3 - 3p^2 = 27
p^2(p - 3p) = 27

anonymous · Answer

sorry second should be p^2(p - 3) = 27

anonymous · Answer

Kind of. But then you're going to have trouble solving that for p. It's easy if you have 0 on the right.

anonymous · Answer

The stuff about first derivative: that's just me confusing myself. Ignore that, I think.

anonymous · Answer

but what would you do with the 27 if you want a zero there?

anonymous · Answer

This can't really be factored. According to wolfram alpha, this has one root, and it looks like an irrational number, which means you can't really write it (unless you invent some symbols). http://www.wolframalpha.com/input/?i=p^3+-+3p^2+-+27

anonymous · Answer

Well, you usually wouldn't move the 27 over to the right, so that you always have a zero on the right.

For example, if you wanted to solve x^2 + 5x + 6 = 0 then you can write x^2 +5x + 6 = (x + 2) (x+3) = 0.

So you never moved the 6 to the right. You just included that in the factorization. You've moved the 27 to the right because you couldn't figure out a way to include it in your factorization. But that means you have problems later, and so it wasn't really that helpful.

anonymous · Answer

That website I linked to has some quite good algorithms. With a problem like this that's quite simple, you can probably be sure that if it could factorize it, then it would have. Instead it gave an irrational solution. So, you just can't factorize this...

anonymous · Answer

Where has this problem come from?

anonymous · Answer

It's a problem created by a teacher.