Help with quartic equation

debjit625 · 2012-07-23 05:56:46

Hi its my first post in this forum...

The question is from a complex number problem I was doing ... but at a point I could not understood how it has been done.

Question : if

,find the value of

Solution :

Now,

I didn't understood this part

What I didn't understood was how the quartic equation

was rewritten to ,whats the process to do that.I could see that the quadratic equation has been used as a factor...

Hope I am clear about my problem

bobbym · 2012-07-23 06:03:51

Hi;

Somebody has been using a package!

Or someone had prior information about the factors. Horner's method is the acceptable way to substitute into a polynomial.

anonimnystefy · 2012-07-23 06:15:39

I think you can just do polynomial division here and substitute x in the remainder.

bobbym · 2012-07-23 06:19:38

How would you know the factors beforehand?

anonimnystefy · 2012-07-23 06:28:20

Factors of what?

bobbym · 2012-07-23 06:34:08

The polynomial.

How does the questioner know this without already knowing the factors? Or is he a darn good guesser?

anonimnystefy · 2012-07-23 06:37:44

He doesn't need factors. If he divides the quartic with the minimal polynomial whose root is the given x, he gets some remainder. He substitues the x into the remainder and calculates it. It should be noted that the remainder is not going to be of degree greater than one.

bobbym · 2012-07-23 06:41:52

What root? The roots are not given to you?

anonimnystefy · 2012-07-23 06:46:10

He constructed a polynomial with one root of -5+4i, didn't he?

bobbym · 2012-07-23 06:49:50

Yes, he did but the problem does not say that is a root. It just says x = something. To plug in x can equal anything. That is what I have been trying to get at.

Bob · 2012-07-23 06:58:34

hi debjit625

Welcome to the forum!

Hi all,

Can you not force this line :

as follows:

Extract the x^2 at the start.

Fiddle the next term so the x^3 comes out OK.

Fiddle the next term so the x^2 comes out ok.

As it happens the x term works too, but if it doesn't, just substitute in the given value of x and it's done.

You don't have to use a package.

Make up a similar one and I'll prove it.

Bob

bobbym · 2012-07-23 07:03:09

Hi;

Yes but he can just hornerize everything and plug in 4i-5

This yields fewer computations.

anonimnystefy · 2012-07-23 07:10:48

But he doesn't want to do it through computation. This way he can know what parts of the quartic are equal to zero. Could you do something for me? Divide the quartic by the quadratic he got. Post your answer in form P(x)=Q(x)*A(x)+R(x), where P is the quartic, and Q is the quadratic x**2+10x+41.

bobbym · 2012-07-23 07:15:18

Hi;

Ask me for my gf's phone number, or 2 pints of blood or the location of my bankroll or the location of the secret base in the desert but please do not ask me that. I beg you.

anonimnystefy · 2012-07-23 07:16:27

Why not. Just divide them. I think M should be able to do that.

bobbym · 2012-07-23 07:18:50

Yes, I am sure it could.

Bob · 2012-07-23 07:20:21

You (or maybe it has to be me) can do that without Mathematica. That's what I'm saying.

Do you want me to?

Bob

anonimnystefy · 2012-07-23 07:25:11

Who was that to?

Bob · 2012-07-23 07:27:40

You and bobbym,

Too late, I'm half way through doing it anyway.

Bob

anonimnystefy · 2012-07-23 07:29:59

Doing what?

Bob · 2012-07-23 07:36:19

This:

So the x^4 term is correct.
But I need 9x^3 not 10x^3

So subtract one x^3

So now the x^3 term is correct.

But I need 35x^2 and I have 41 10 = 31.

So I need another 4x^2

The x^2 term now is correct.

The x term is -41x + 40x = -x

That part was just lucky. It is correct too.

So

Bob

anonimnystefy · 2012-07-23 07:58:07

And that's it. No finding roots at all!

debjit625 · 2012-07-23 19:04:36

First of all thanks to everybody for your replies..

@bob bundy
Thanks and I think "anonimnystefy" mentioned the same process.

I thought of this process earlier as its already solved and I just needed to rearrange it, but it was too lengthy and I thought their might be a simpler way to do it...but now I thing I might be wrong.

If anybody knows any easier process please let me(us) know. I will keep an eye on this post...

Anyway thanks again

anonimnystefy · 2012-07-23 21:31:56

anonimnystefy wrote:

He doesn't need factors. If he divides the quartic with the minimal polynomial whose root is the given x, he gets some remainder. He substitues the x into the remainder and calculates it. It should be noted that the remainder is not going to be of degree greater than one.

Hi

Here is what I proposed, but I do not if Bob also got it or did by mine, but I think it is the former.

Either way, you could just substitute directly, but calculaing that would probably be more lengthy than the process Bob showed here.

Math Is Fun Forum

#1 2012-07-23 05:56:46

Help with quartic equation

#2 2012-07-23 06:03:51

Re: Help with quartic equation

#3 2012-07-23 06:15:39

Re: Help with quartic equation

#4 2012-07-23 06:19:38

Re: Help with quartic equation

#5 2012-07-23 06:28:20

Re: Help with quartic equation

#6 2012-07-23 06:34:08

Re: Help with quartic equation

#7 2012-07-23 06:37:44

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#8 2012-07-23 06:41:52

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#9 2012-07-23 06:46:10

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#10 2012-07-23 06:49:50

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#13 2012-07-23 07:10:48

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