Home » Ask & Discuss » Mathematics. » Algebra « Back to Discussion

Algebra

Forum Expert

Joined:	28 Feb 2007
Post:	2173

3 Mar 2008 08:43:22 IST

0 People liked this

799

Polynomial value

None

f(x) = x³+x+1. g(x) is a cubic polynomial such that g(0) = -1 and the roots of g are the squares of the roots of f. find g(9).

Share this article on:

Comments (8)

anchit saini

Blazing goIITian

Joined: 1 Feb 2008

Posts: 1251

3 Mar 2008 08:56:58 IST

Like 0 people liked this

899 ??

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

3 Mar 2008 09:12:16 IST

Like 0 people liked this

approach pls?

anchit saini

Blazing goIITian

Joined: 1 Feb 2008

Posts: 1251

3 Mar 2008 09:13:52 IST

Like 0 people liked this

sir is it correct?

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

3 Mar 2008 09:17:18 IST

Like 0 people liked this

yeah, answer is right. i want to see how you did it.

anchit saini

Blazing goIITian

Joined: 1 Feb 2008

Posts: 1251

3 Mar 2008 09:19:59 IST

Like 3 people liked this

for f(x)---
roots be a, b, c

a+b+c=0          ------1
ab+bc+ca=1    -------2
abc=-1            ------3

g(x)=px^3 + qx^2 + rx -1

roots are
a^2, b^2, c^2

a^2*b^2*c^2=1/p
also from eqn 3, abc=1

hence1/p=1
p=1

a^2+b^2+c^2=-q
also
(a+b+c)^2=-q + 2(ab+bc+ca)
0=-q+2 from eqn1 and 2
hence
q=2

a^2b^2 +b^2c^2 + a^2c^2=r
and
also
(ab+bc+ca)^2=r + 2abc(a+b+c)=r+0        from eqn 1

1=r         from eqn 2

hence
g(x)=x^3+ 2x^2 + x - 1

now we can put x=9 to get the answer

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

3 Mar 2008 09:21:27 IST

Like 0 people liked this

there are shorter approaches. try and find them

Sairam

Blazing goIITian

Joined: 14 Sep 2007

Posts: 573

3 Mar 2008 14:14:23 IST

Like 4 people liked this

f=x^3+x+1 let its roots be a,b,c

den f= (x-a)(x-b)(x-c)

Now g(x)= a(x-a^2)(x-b^2)(x-c^2)

g(0)=-1--->a=1

g(9)=(3+a)(3+b)(3+c)(3-a)(3-b)(3-c)

=f(3)*f(-3)*-1

=31*29

=899

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

3 Mar 2008 14:24:47 IST

Like 3 people liked this

lovely elasti, that's how I had worked it out.

But, here's another nice method, and I think Feynmann in this forum has used earlier.

x³+x+1 = 0

x(x²+1) = -1

x²(x²+1)² = 1

If x² = y, then y(y+1)² =1

or y³+2y²+y-1 = 0.

The roots to this cubic are the squares of the roots of x³+x+1 = 0

Hence g(y)

y³+2y²+y-1 as it satisfies g(0) = -1

Hence g(9) = 899.

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team