Math Is Fun Forum

rickyoswaldiow,

It looks to me as though you're saying the question is:

f(x) = 3x^4 + ax^3 + bx^2 - 7x - 4, find a and b such that f(x) has (x-1) as a factor. Are you sure that is what the question is asking, because I don't think that's enough information to solve it.

In general, I'd say you can't pin down two unknowns from one factor. When you work through the polynomial division, you need to end up with a remainder of zero at the end, but your equation for the remainder will be in terms of both a and b. And one equation just isn't enough to solve for two unknowns.

Errr ...

question 2 is really going to bother me for a while I think

Now I see what you were saying. When solving algebra equations, if you apply a square root to both sides it is then necessary to be careful to choose the correct + or - sign when you apply it, because the correct solution to your equation might only be valid in one of those cases.

For this problem, it really doesn't matter which square root you do in line 2 (+sqrt) or (-sqrt) as long as you do the same one to both sides because you aren't trying to solve for a variable from line 1. Line 2 is a fully correct statement by itself: the sqaure root of -1 does equal the square root of -1. It probably would have been clearer if I had just started the proof from line 2.

It is the second mistake that Ricky found that I claim is the real problem with this proof. You can't go from root(1/-1) to root (1)/root(-1). The first equals i, and the second equals i raised to the minus one power. The cyclical property of imaginary numbers was one thing I was hoping to get to here:
.
.
i^0 = 1
i^1 = i
i^2 = -1
i^3 = -i
i^4 = 1
i^5 = i
.
.
.
And so on in both directions. So by splitting the square root on the right side in line 4, you are effectively saying i^1 = i^-1, which is not true, it's two off on the sequence.

I assume you mean a second degree polynomial equation here? There are two (though once in a while they are the same as each other). I think I see where you're going with this, but it is also quite possible to find the problem with this proof using only what we know about the properties of complex numbers. These are properties that I believe are taught in the second year of Algebra at most high schools.

I'm not 100% sure that I understand the rules in your question, but it sounds to me as though your concern is that on an infinite axis, you can't tell the robots to go in a given direction because if you pick the wrong direction then they'll never hit an end and so they'll never know that they're getting farther apart.

A recommendation for how you might get around this: do a scan. As an example, take a step to the right, check for a flag. If you don't find one, take two steps back to the left, check for a flag. If you don't find one, take three steps back to the right, and so on, going out in both directions from the center. If you're goal is to maximize efficiency then I'm sure this isn't the fastest way, but something along those lines would work. You just have to make sure you don't take such big steps that one robot might miss the flags left by the other. You then have to deal with what to do when one robot finds a flag left by the other, but I hope something like this could at least get you started.

Ah, I did not look at the sticky at the top of the page. Yes, LaTeX is nice.