View Full Version : Roots of equation

jake2954

07-25-2011, 04:03 PM

The equation X^4 - 3X^3 + 19X^2 +53x - 174 has four roots. If two of the roots are real and equal to x=-1 and x=3 , by a process of long division and solving a quadratic equation, find the two complex roots.

Please can you help me with this question as I am struggling to anwer it. :(

Thanking you in advance

Jake

JeffM

07-25-2011, 05:17 PM

The equation X^4 - 3X^3 + 19X^2 +53x - 174 has four roots. If two of the roots are real and equal to x=-1 and x=3 , by a process of long division and solving a quadratic equation, find the two complex roots.

Please can you help me with this question as I am struggling to anwer it. :(

Thanking you in advance

Jake

Hi Jake

EDIT: ARE YOU SURE YOU WROTE BOTH THE QUARTIC AND THE ROOTS CORRECTLY?

(-1)[sup:1r78g2ov]4[/sup:1r78g2ov] - 3(-1)[sup:1r78g2ov]3[/sup:1r78g2ov] + 19(-1)[sup:1r78g2ov]2[/sup:1r78g2ov] + 53(-1) - 174 = 1 + 3 + 19 - 53 - 174 = 23 - 227 = - 204, which is not quite zero.

I give help but do not solve your problems for you. Most people here do the same.

Did you know that if P(x) is a polynomial of degree n and P(a) = 0, there exists Q(x) such that (x - a) * Q(x) = P(x) and Q(x) is a polynomial of degree (n - 1)?

This is a very important theorem based on the Fundamental Theorem of Algebra.

So what is the relevance of this theorem to your problem?

It means that x[sup:1r78g2ov]4[/sup:1r78g2ov] - 3x[sup:1r78g2ov]3[/sup:1r78g2ov] + 19x[sup:1r78g2ov]2[/sup:1r78g2ov] + 53x - 174 = (x + 1) * (x - 3) * R(x) = (x[sup:1r78g2ov]2[/sup:1r78g2ov] - 2x - 3) * R(x), where R(x) is a quadratic. Do you see why?

OK. So that means R(x) = (x[sup:1r78g2ov]4[/sup:1r78g2ov] - 3x[sup:1r78g2ov]3[/sup:1r78g2ov] + 19x[sup:1r78g2ov]2[/sup:1r78g2ov] + 53x - 174) / (x[sup:1r78g2ov]2[/sup:1r78g2ov] - 2x - 3).

Do you know how to do algebraic long division?

Denis

07-25-2011, 08:50 PM

The equation X^4 - 3X^3 + 19X^2 +53x - 174 has four roots. If two of the roots are real and equal to x=-1 and x=3 .....

That's not an equation; this is: X^4 - 3X^3 + 19X^2 +53x - 174 = 0

As Jeff told you, those x values are WRONG; where/how did you get them?

Should be x = -3 and x = 2.

NOTE: do not use X and x in same equation...capish?

jake2954

07-26-2011, 04:26 AM

The equation X^4 - 3X^3 + 19X^2 +53x - 174 has four roots. If two of the roots are real and equal to x=-1 and x=3 , by a process of long division and solving a quadratic equation, find the two complex roots.

Please can you help me with this question as I am struggling to anwer it. :(

Thanking you in advance

Jake

Hi Jake

EDIT: ARE YOU SURE YOU WROTE BOTH THE QUARTIC AND THE ROOTS CORRECTLY?

(-1)[sup:32sqivez]4[/sup:32sqivez] - 3(-1)[sup:32sqivez]3[/sup:32sqivez] + 19(-1)[sup:32sqivez]2[/sup:32sqivez] + 53(-1) - 174 = 1 + 3 + 19 - 53 - 174 = 23 - 227 = - 204, which is not quite zero.

I give help but do not solve your problems for you. Most people here do the same.

Did you know that if P(x) is a polynomial of degree n and P(a) = 0, there exists Q(x) such that (x - a) * Q(x) = P(x) and Q(x) is a polynomial of degree (n - 1)?

This is a very important theorem based on the Fundamental Theorem of Algebra.

So what is the relevance of this theorem to your problem?

It means that x[sup:32sqivez]4[/sup:32sqivez] - 3x[sup:32sqivez]3[/sup:32sqivez] + 19x[sup:32sqivez]2[/sup:32sqivez] + 53x - 174 = (x + 1) * (x - 3) * R(x) = (x[sup:32sqivez]2[/sup:32sqivez] - 2x - 3) * R(x), where R(x) is a quadratic. Do you see why?

OK. So that means R(x) = (x[sup:32sqivez]4[/sup:32sqivez] - 3x[sup:32sqivez]3[/sup:32sqivez] + 19x[sup:32sqivez]2[/sup:32sqivez] + 53x - 174) / (x[sup:32sqivez]2[/sup:32sqivez] - 2x - 3).

Do you know how to do algebraic long division?

Hi

Yes I have got that far and with alegraic long division i get x[sup:32sqivez]2[/sup:32sqivez]-x+20+(20x+114/x[sup:32sqivez]2[/sup:32sqivez]-2x-3)

Just not sure where to go beyond there as i have a remainder.

Thanks everyone for your contributions so far.

Subhotosh Khan

07-26-2011, 10:32 AM

The roots that you wrote are wrong.

Denis gave you the correct roots.

You should divide the polynomial by (x+3)*(x-2) [= x[sup:3gngpol1]2[/sup:3gngpol1] + x - 6] ? then you would not have any remainder.

.

jake2954

07-26-2011, 11:48 AM

The roots are quoted in the question.

Is there a way to proceed with the quoted roots, which has a remainder?

Or is the question impossible to answer with the parameters given.(There may be a possibility that the question was cut and pasted)

JeffM

07-26-2011, 03:20 PM

The roots are quoted in the question.

Is there a way to proceed with the quoted roots, which has a remainder?

Or is the question impossible to answer with the parameters given.(There may be a possibility that the question was cut and pasted)

As I said in my original post, perhaps the roots are OK but the polynomial is misstated. You need to check both. If both polynomial and roots are as stated, the problem is insoluble because those values are not roots of that polynomial. If they were there would be no remainder

Powered by vBulletin® Version 4.2.2 Copyright © 2014 vBulletin Solutions, Inc. All rights reserved.