Hi there,
I'm having a hard time finding the roots of this equation:
I've been trying to figure out higher degree polynomials, quadratics are fine, but anything above that (cubic etc) I just can't understand!
Hope you can help! Thanks!

Hi there,
I'm having a hard time finding the roots of this equation:
I've been trying to figure out higher degree polynomials, quadratics are fine, but anything above that (cubic etc) I just can't understand!
Hope you can help! Thanks!
How about letting and treating it as a quadratic in ?
Hi there, thanks for the reply!
You mean like this?
Where
I'm a little confused, sorry. Hope you can explain a bit more?
Thanks!
No, JaneBennet means that you may substitute all the x^2 by y.
This will make your equation look like that:
y^2  3y + 2 =0
which you can solve. Then, having the 2 values of y, you get the 4 values of x.
This will always work here as long as you don't have the disturbing terms x or x^3. If that happens, you can't use that method.
I don't think there is a general easy way to get the roots. There are of course general cubic and quartic formulae, but they are very "scary".
For your equation, the first idea that came to me was this:
x^4  3 x^2 + 2 = 0 implies
x^2(x^2  1)  2(x^2  1) = 0 implies
(x^2  2)(x^2  1) = 0 implies
x^2 = 2 or 1 implies
x = 1, 1, sqrt2, sqrt2
This "separation method" is my favourite. But I don't suggest trying that in an exam as it may take time to figure out how to separate your terms.
So I think the most efficient method would be to guess a root. Try x = 1. When you put it in your equation, you get 0, so x = 1 is a root. Hence, you may divide your equation by (x1). You get a cubic equation. Now try something else. For example x = 1. Good, another zero. Hence, divide by (x+1). you get a quadratic equation, which is easy. This method is longer but more methodic...
You may improve that method by guessing 2 roots: for instance try x = 1. You get zero. Don't divide yet: try x = 1. Again zero. So you may divide your equation directly by (x^2 1) which is nothing but (x1)(x+1)...
But of course, if your equation looks nice like here, just make the substitution of JaneBennet.
oh, and I forgot to mention a method that will greatly reduce the number of guesses. By elementary arithmetic, your "guess" must divide the "free" term (is it how you call it?). By the "free" term I mean the term with no x, which here is 2. (sorry if I'm not using the correct name). So 1, 1, 2, 2 are good guesses. (the guess x=5 for instance will never work).
2,2 won't work, but 1,1 will.
By the way, the general formula can be found here:
http://planetmath.org/encyclopedia/QuarticFormula.html
but it's not very practical...
Starting from degree 5 and above, you have no formulae at all... but to prove this you need some good math... very beautiful math by the way, I think one of the problems of math taught at school is that most of it is really boring. The math you get at the university is really something else, infinitely more beautiful and interesting...
Hi Stranger,
Thanks for the reply! And apologies for my late reply. Thanks, that's very helpful, will have a good look!
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