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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -




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Topic review (newest first)

2012-09-27 18:12:32


Clever and compact.

2012-09-27 12:48:05

Hi again bobbym!

I couldn't agree more that math should be neat, clean and simple enough for everyone to grasp.
That's been my passion and my focus most of my career --- viewing math as a language and
trying to see if the symbolism, definitions, algorithms, etc. can be simplified and improved.  It
really needs close scrutiny since it evolved over many centuries by folks that had no chance to
communicate with each other to try to make it really consistent, coordinated, correct, concise, and
any other word we can think of that starts with a "c".


2012-09-27 09:23:16

Hi noelevans;

It is nifty. Neat and clean and simple enough for everyone to grasp. That is what math should be like.

2012-09-27 08:11:25

Hi bobbym!

I've never seen the generalized "sum of roots" formula you gave in post#6.  That's really nifty!
Thanks for sharing it.  smile

2012-09-25 22:06:12


You can use the theory of equations to try another way. It says there is a relationship between the roots and the coefficients.

The sum of the roots of a nth degree polynomial

are equal to

So you have this equation to solve

solving for r3 you get r3 = -1 which is the third root.

bob bundy
2012-09-25 21:18:48

Well, as you know that cubic = linear x quadratic you can sort of figure out the quadratic coefficients as you go.  It takes less time but amounts to short cut division so it's not really a new method. 

Also you could call the quadratic coefficients a, b and c and do a bit of algebra to get them, but it's still pretty much the same.

In short, I think you have the optimum method already.


2012-09-25 20:20:17

So, Is division the only way?

bob bundy
2012-09-25 20:01:08

hi Agnishom and Au101

That approach looks like the most straight forward to me.

For this particular problem

you might notice that

which means you know (x+1) is factor straight away by the factor theorem.

In general, dividing by known factors is the way.


ps.  For typical exam questions, they cannot choose factors that would take a long time to find, so I always do a quick mental check for x = +/-1, +/-2, +/-3.  If I haven't found a factor by then  I do another quicker question first.

2012-09-25 19:43:15

I'm not sure what exactly your approach would be. Originally, I just posted my approach:

If 2 is a zero of the polynomial, then (x - 2) is a factor.

(By polynomial long-division)

Therefore, the other zero is -1

But having re-read your first post, I think that might be what you would have done anyway. I'm not sure that I know any more efficient method. But I suggest you just divide once and then factorise, that - at least - might make things a little faster?

2012-09-25 18:15:19

The two zeros of a polynomial are


and the polynomial is

What is the third zero?

My approach to such kinds of problem is to divide the polynomial by the two given factors to obtain the other factor and then the zero.

Can Someone suggest a quicker and more efficient method?

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