Factoring (I think?)

I'm just starting a calculus course and it has been ages since I've done anything algebra-related.

I've ran into some issues and I don't even know where to start. I think it is factoring, but none of the sites have given me enough information on what I need to do.

I am hoping someone could help me out.

So I have this equation (including everything, but I don't think the limit is needed..correct me if I'm wrong).

lim h->3

h^3-27
----------
h-3

Is the equation I'm working with. That is h cubed minus 27 over h minus 3.

Now I've looked at the answer and it has step by step how to get to the answer, but the first step is what has me confused.

My question is how does the above equation turn into:

(h-3)(h^2+3h+9)
-----------------------
h-3

Any information at all would help. Key terms or a site I could review to get the hang of this again would be best.

You are right, it is factoring. All they did was factor h³-27 into (h-3)(h²+3h+9) so the factor (h-3) cancels out of the numerator and denominator.

The factoring can be done by dividing h³-27 by (h-3) using what's called synthetic division. See below link for a step by step example.

http://tutorial.math.lamar.edu/Class...lynomials.aspx

Small thing: the x's in the above post should be h's.

(Fixed; delete this post if someone with the ability to do so comes around)

Quote:

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Originally Posted by jemidiah

Small thing: the x's in the above post should be h's.

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Oops, I fixed it.

Quote:

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Originally Posted by jemidiah

Small thing: the x's in the above post should be h's.

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Oops, I fixed it. Thanks.

Thanks for the help on the first one, I was also told by a co-worker of the rule of cubes, and so I was able to get that one using that.

There is another one I don't get either... I've read through the page you listed, but that is all finding how many times the denominator can go into the numerator. Maybe I don't understand this, but as far as I can tell with this one, the denominator cannot go into the numerator.

So it says to write out the exponents in order. So, x^2 - 1 over 3x+3 should turn into

x^2 + 0x -1 over 3x + 3, correct?

But 3x + 3 doesn't go into x^2 + 0x.

So I think it turns into (x-1)(x+1) over 3x+3, but I'm stuck here again...

It's right in front of you. :)

3x+3 = 3(x+1)

Could you explain how you got to that? The book has the answer as negative 2/3 (the whole fraction is negative).

*edit* I forgot to add that the above is a limit as well. x->-1

(x-1)(x+1)/(3x+3) = (x-1)(x+1)/(3(x+1)) = (x-1)/3

The (x+1) in the numerator and denominator cancel.

Quote:

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Originally Posted by kfcSmitty

Maybe I don't understand this, but as far as I can tell with this one, the denominator cannot go into the numerator.

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You probably tried plugging -1 into the numerator and denominator at first, and got 0/0. Since the denominator has only an x term (not x^2, or x^3, or higher) and the numerator shares a root, the denominator *must* go into the numerator.

More generally, say you have a numerator N(x) that's a polynomial, and a denominator (ax-b) for some values of a and b. This still fits the situation you gave--N(x) is just x^2 - 1, and since the denominator is 3x+3, a=3 and b=-3. In this case, suppose N(L) = 0 for some value L that also makes ax-b = 0. L = -1 in your example, which continues to fit, since (-1)^2 - 1 = 0, and 3(-1) + 3 = 0. Since N(x) = 0 when x = L, it turns out (x-L) must be a factor of N(x). That is, N(x) = P(x) * (x-L) for some lower degree polynomial P(x). In your example, P(x) ends up being (x-1). Further, the (x-L) term is just a scale factor from being equal to (ax-b). Why? We know aL-b = 0, since ax-b = 0 when x = L. So, aL = b, and L = b/a. The (x-L) factor is then (x-b/a), which is 1/a*(ax-b). Putting it together, you have a string of equalities:

N(x) / (ax-b) = P(x) * (x-L) / (ax-b) = P(x) * (x-b/a) / (ax-b) = P(x) * 1/a * (ax-b) / (ax-b) = P(x) / a

or, plugging in values from your problem,

(x^2 - 1) / (3x-(-3)) = (x-1) * (x-(-1)) / (3x-(-3)) = (x-1) * (x-(-3/3)) / (3x-(-3)) = (x-1) * 1/3 * (3x-(-3)) / (3x-(-3)) = (x-1) / 3, as VBAhack said

So in this general situation, it'll always simplify, somehow. The important thing to get out of this (since I'm pretty sure you'll get at most the high points of the above paragraph; not tryin' to be mean, just realistic) is to think about any polynomial you have as a product of linear factors. All you need to know are the roots of the polynomial--where it's zero--and you'll know all of those linear factors exactly. From there, you can simplify them or classify them more easily.