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#1
11-16-2007, 09:50 AM
 Retrictory_nomeal Junior Member Join Date: Nov 2007 Posts: 1
Polynominal problem

I'm finding trouble to solve the following question:

Find the cubic equation whose roots are the sqaures of that of
x^3 + 2x +1=0.

Any help would be greatful
#2
11-16-2007, 02:52 PM
 HallsofIvy Senior Member Join Date: Oct 2006 Posts: 464

If a, b, c are roots of a cubic then (x-a)(x-b)(x-c)= 0.
Multiplying that out, x^3-(a+ b+ c)x^2+ (ab+ac+bc)x- abc= 0
Compare that to x^3+ 2x+ 1= 0. a+b+c= 0, ab+ac+bc= 2, abc= -1.

If a^2, b^2, c^2 are the roots, then you have
x^3-(a^2+b^2+c^2)x^2+ (a^2b^2+a^2c^2+ b^2c^2)x- a^2b^2c^2= 0.
Knowing a+b+c= 0, ab+ac+bc= 2, abc= -1, how can you find those coefficients?
(Looking at (a+b+c)^2 might help.)

Last edited by HallsofIvy : 12-04-2007 at 01:31 PM.
#3
11-21-2007, 05:18 PM
 krusta80 Junior Member Join Date: Nov 2007 Posts: 1

sosmath.com/algebra/factor/fac11/fac11.html

Wish I could take credit for it, but oh well.

I'm looking forward to solving cubic equations now without all of the guesswork.
#4
11-26-2007, 11:36 AM
 HallsofIvy Senior Member Join Date: Oct 2006 Posts: 464

Yes, the cubic formula is nice but I don't see what that has to do with this problem.
#5
11-29-2007, 01:51 AM
 Leon323 Junior Member Join Date: Nov 2007 Posts: 2

hmm the cubic formula is too long man maybe here he can get more help
__________________
http://thatmathsite.googlepages.com/home <<<CLICK THERE FOR MORE HELP PLEASE
#6
12-04-2007, 01:39 PM
 HallsofIvy Senior Member Join Date: Oct 2006 Posts: 464

Time to put you out of your misery! I said before
Quote:
 Originally Posted by HallsofIvy If a, b, c are roots of a cubic then (x-a)(x-b)(x-c)= 0. Multiplying that out, x^3-(a+ b+ c)x^2+ (ab+ac+bc)x- abc= 0 Compare that to x^3+ 2x+ 1= 0. a+b+c= 0, ab+ac+bc= 2, abc= -1. If a^2, b^2, c^2 are the roots, then you have x^3-(a^2+b^2+c^2)x^2+ (a^2b^2+a^2c^2+ b^2c^2)x- a^2b^2c^2= 0. Knowing a+b+c= 0, ab+ac+bc= 2, abc= -1, how can you find those coefficients? (Looking at (a+b+c)^2 might help.)
(a+ b+ c)^2= a^2+ b^2+ c^2+ 2ab+ 2ac+ 2bc.
We know from above that a+ b+ c= 0 so (a+ b+ c)^2= 0. We know also that ab+ ac+ bc= 2 so 2ab+ 2ac+ 2bc= 2(ab+ac+ bc)= 4. That is,
a^2+ b^2+ c^2+ 4= 0 so a^2+ b^2+ c^2= -4.
a^2b^2c^2= (abc)^2 and abc= -1 so a^2b^2c^2= (-1)^2= 1.

So far, the polynomial must be
x^3+ 4x^2+ (a^2b^2+ a^2c^2+ b^2c^2)x- 1.

To find a^2b^2+ a^2c^2+ b^2c^2, look at (ab+ac+bc)^2= a^2b^2+ a^2c^2+ b^2c^2+ 2a^2bc+ 2ab^2c+ 2abc^2.
We know that ab+ac+bc= 2 so (ab+ac+bc)^2= 4.
2a^2bc+ 2ab^2c+ 2abc^2= 2abc(a+ b+ c). We know that abc= -1 and that a+ b+ c= 0 so a^2b^2+ a^2c^2+ b^2c^2+ 2(-1)(0)= 4 which tells us that a^2b^2+ a^2c^2+ b^2c^2= 4.

If a, b, c are the roots of the cubic equation x^3+ 2x+ 1= 0, then x^3+ 4x^2+ 4x- 1= 0 has roots a^2, b^2, and c^2.

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