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The Autobiography of Russell
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A little math help please?
If there's anyone out there who can help me check over a math problem I would be very appreciative.

Now, before I get started I just want to make it clear that this is a problem on a take-home quiz. Just making that clear in case someone has some objections over it.

The problem is to Evaluate f(x) = -x^2 - 2x - 3 where f(x+h). This is what I've gotten so far:
-(x+h)^2 - 2(x+h) - 3
-(x^2 + h^2 + 2xh) - 2x - 2h - 3
-x^2 - h^2 - 2xh - 2x - 2h - 3

Well, that is as far as I can think to get. Am I missing something here or is that the answer?

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3 comments or Leave a comment
raist_ From: raist_ Date: February 20th, 2006 07:26 pm (UTC) (Link)
Looks right to me. You might be able to factor it down, but in the space of a few seconds, I can't see how or where.
zimzat From: zimzat Date: February 20th, 2006 08:43 pm (UTC) (Link)

Yeah, one of the problems I saw later that involved h did something like that, except it's not so clear cut here.

-x^2 - h^2 - 2xh - 2x - 2h - 3
-x^2 - 2x - h^2 - 2h - 2xh - 3
-x(x + 2) - h(h + 2) - 2xh - 3
(I left 2xh out of either one because it has both x and h in it and putting it in one would put the other in it as well)
raist_ From: raist_ Date: February 20th, 2006 09:19 pm (UTC) (Link)
That 3rd step is usually not done that way.

Generally, when you get terms like that, you're looking to factor into binomials. -x^2 - 2xh - 2h^2 becomes -(x+h)(x+h). Most profs prefer it that way, especially when division is involved, because you have a higher chance of canceling out whole quanitites then just an x or an h.

Factoring and Simplifying really is trial and error, for the most part.
3 comments or Leave a comment