I'm not sure how a more technical proof would go, but basically, any multiple root is going to be at a point where the polynomial touches, but does not cross, the x-axis. (Ok, so that's not always true.) Then the polynomial would have to be flat, and have a slope of zero at that point. Or did you already know all that?

Not taking a math class for the past year has not been good for me.

Haha, yeah I know that stuff.

It's just.. Some sort of proof involved that I must have missed during class.

Oh, never mind then. I have a feeling that I probably proved this at one point, but I don't remember how.

Wait, I might have something. If you express a polynomial with multiple roots as product of several terms of (x-c)ⁿ, where c is the root and n is the multiplicity, like x²(x-3)², you can just write out the general form of such a polynomial. Then you can take the derivative, and show that it will always have the same roots, because the derivative will contain the term n(x-c)^n-1 multiplied by a bunch of other stuff. It might get a little messy, because of the product rule, but it's doable.

I hope you're working with real roots only?

Uhh, I found it eventually in my textbook..

Here it is, if anyone else wants to know:

[Since if P'(a) would mean a double root, we'll try to prove that P'(a) = 0
Since P(x) = 0 has a double root at x = a, and can therefore be written as

P(x) = (x-a)²Q(x)

Therefore,

P'(x) = (x-a)²Q'(x) + Q(x).2(x-a)
P'(x) = (x-a)[(x-a)Q'(x) + 2Q(x)]

P'(a) = (a-a)[(a-a)Q'(a) + 2Q(a)]
P'(a) = 0 x 2Q(a)
P'(a) = 0

As required.

...

Or you could just do that. I guess I wasted all that time editing my post while I was finding the answer.

But hey, at least I can feel good about the fact that my mathematical skills still exist. Also, my version works in a more general case.

:(

Sorry. I didn't see your edit.

It was too late anyway. But I'm glad you figured it out.

I really really really need help on multiplying rational expressions of all kinds (binomials and trinomials in all numerators and divisors) and simplifying them. Help me please?

Give an example, and I'll help.

A few examples would be these:

7 over 6p^2+q divided by 14 over 18p^2+3q

6x+6 over 5 divided by 3x+3 over 10

Edit: I didnt give a trinomial example, heres one

x^2-25 over 2x-2 divided by x^2+10x+25 over x^2+4x-5

That helped me a lot, but what if in the trinomial, say the term usually containing x^2 (forgot the name of that term), has an exponent of 3 or a whole number multiplied to it bigger than 1?

How do you mean? I think in the stuff you're working with, it should all be factorable, still. If it's not, you can just leave it as it is, and it ends up as part of your answer.

nevermind, I figured that part out. But now im having trouble with a fraction. 3x over 6 does not reduce does it? would it be x over 2? Because I get completely different answers with both. Sometimes when you do more complicated math, you forget the easiest I guess

It does reduce, but you should still be getting the same answer with both. It's prolly better to reduce before going on, it simplifies things.

You're right *forehead smack* DURRRR

Shush, you.

*Criticises her partner's choice of paper, but stares happily at his handwriting.*

NEVERMIND. Stupid me.

Okay so I have a set S containing two elements and a set T containing 3 elements and f:S->T. Since this is one-to-one an inverse can be defined right?

No. That seems to be the definition of not one-to-one.

Here are some arbitrary values plugged in...

S: { a, b, c }
T: { x, y }

f(a) -> x
f(b) -> y
f(c) -> x

f^-1(x) = THE UNIVERSE EXPLODES OH GOD WHAT HAVE YOU DONE.

No but I'm mapping from 2 elements to 3 elements. S={1,2} T={A,B,C}.
So, f(1)=A
f(2)=B

Oh, I read that wrong. You're right. Unless you mean any element from T as part of the domain. Then it cannot.

I'm not sure what you intended by "since" in "Since this is one-to-one". If you meant that the mapping is necessarily 1-1, then think again. You could have f(1)=A, f(2)=A.

If you meant "I am also specifying" a 1-1 relationship, then you're nearly there. There will be an inverse, but it only defined over the range of f, not the whole of T. In your example, it is only defined for {A,B}. Since nothing maps to C, there is no inverse available.

(i.e. what Blake said, just spelling it out a bit more.)