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Good to hear -- welcome to the forum!

Draw a Venn diagram with 3 overlapping circles, labelling the four bits that overlap as A, B, C, D.

The information given in the question then allows you to form some equations for A, B, C and D. Can you see how?

!nval!d_us3rnam3 wrote:

That depends on what you consider to be a 'rigorous demonstration'. Clearly is a factor, since both the LHS and RHS vanish. Dividing your equation by (taking ) gives youSince vanishes whenever , then the numerator of the term on the RHS also vanishes for these values of . That's three solutions, and you can 'rigorously demonstrate' that there aren't any more solutions over via the method in post #4.
Ok, but I wasn't looking for a proof with the answers in the beginning. I've gotten pretty far, but I need to find all the answers to

EDIT: With a rigorous demonstration that this is the answer.

Hi Stuti55,

Welcome to the forum.

Hi Zeeshan,

I've moved your thread to the Help Me ! forum.

Have you checked out **Maths Is Fun's page on long division**?

Hi Anduin,

Welcome to the forum. What features does the graph y = sin(x) have? How do they compare to the graphs y = 2sin(x) and y = 3sin(x)?

Hello,

I noticed that you've posted several times here as a guest. Why not register an account with us?

What have you tried? What sorts of restrictions do you think you can place on x and y?

Grantingriver wrote:

Hi Zetafunc, if there are duplicate roots of a polymolial you should counts them as distinct roots.

Why? Precisely the opposite is true: if a root is repeated, then we cannot call those repeated roots 'distinct', and whether or not we count them in the same way we would distinct roots depends on the context.

Grantingriver wrote:

For example, if you say that a give polynomial has “exactly” the roots -3, 4 and 8 that means there is no duplicate roots.

I am not convinced that the word 'exactly' necessarily implies distinctness -- I would have thought that the multiplicities were worth consideration, otherwise the problem appears to be a little simplistic.

Grantingriver wrote:

This is not quite true: the multiplicities of the zeroes are important, e.g. and .
The answer for part 1 is, by all means, yes. This is so because the condition that f(x) has the zeros -3,4 and 8 entails that:

f(x)=(x+3)×(x-4)×(x-8)

And since the zeros of g(x) are -5,-3, 2, 4 and 8 we also have:

g(x)=(x+5)×(x+3)×(x-2)×(x-4)×(x-8)

and hence:

g(x)/f(x)=[(x+5)×(x+3)×(x-2)×(x-4)×(x-8)]/[(x+3)×(x-4)×(x-8)]=(x+5)×(x-2)

and since the result of the division (x+5)×(x-2) is a polynomial without a remainer, therefore g(x) is divisible by f(x). The answer of part 2 is very clear from the answer of part 1. For a polynomial g(x) to be divisible by a polynomial f(x) at least all the roots (zeros) of f(x) should be also roots (zeros) of g(x). And the final statement completes the answer.

In particular, the second pair of brackets in the second line and the first pair of brackets in the third line are not correct.

However, the question appears to be badly worded. The area 'bounded' by the curve, the axes and the line x = 2 is not bounded at all -- it is infinite.

Yes, that is correct.

Yes -- do you know how to calculate this?

One way is to use the binomial distribution.

Suppose that represents the number of times that the goalkeeper saves a penalty. Then, has a binomial distribution: You want to calculate . Do you see why?Hi MellyBigD,

Welcome to the forum. You have posted 4 threads asking for help, but you have not shown any attempts at the problems or given any pointers about where you are stuck. We will endeavour to help you as best you can, and we happily volunteer our time to do so, but we will not do your homework for you. Can you please show some attempts at these problems, or be specific about what exactly you are stuck on?

False: consider the equation .

What happens when x is greater than or equal to 1?

What is the limit of your function as x approaches 2?

There are analogous formulae for higher order derivatives too, and several published papers about the error term.

Hi Sara,

Thanks for stopping by -- have you considered registering an account with us?

Have you heard of the chain rule, i.e. ?Monox D. I-Fly wrote:

So, the approximation of x doesn't have to be 0 to make it equals 1?