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Algebra

Cool goIITian

 Joined: 15 Jul 2007 Post: 75
10 Feb 2008 13:31:17 IST
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7
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polynomial
Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Algebra

let  p(x)=x^2+bx+c
where b and c are integer if p(x) is a factor of both x^4+6x^2+25
and 3x^4+4x^2+28x+5 what is p(1)

sorry

Blazing goIITian

Joined: 9 Jan 2008
Posts: 843
10 Feb 2008 13:34:23 IST
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what is the last eqn

Blazing goIITian

Joined: 9 Jan 2008
Posts: 843
10 Feb 2008 13:59:41 IST
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Look at my soln

here

p(x)=x2+bx+c
p(1)=1+b+c

since it is a factor of the given two eqn , itmust surely satisfy them

(b+c+1)4+6(b+c+1)2+25=0...............(1)

3(b+c+1)4+4(b+c+1)2+28(b+c+1)+5=0............(2)

multiply by 3 in eqn (1)

and subtract
u''ll get

14(b+c+1)2-28(b+c+1)+70=0

Put  T=b+c+1

14T2-28T+70=0
T2-2T+5=0

Now here Disc<0 which is not possible ,hence i think  in the first eqn it should be -25 not +25.
if u take -25

u'' get this eqn

T2-2T-5=0

T=1+-root(6)

b+c+1=1+-root(6)
b+c= +-root(6)
p(1)=b+c+1
p(1) = +-root(6)+1

Forum Expert
Joined: 28 Feb 2007
Posts: 2185
10 Feb 2008 14:33:29 IST
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P(1) would have been a root of both equations if 1 is a root of x2+bx+c which is not known be true at this stage.

My attempt:

It is given that x4+6x2+25 = P(x) Q(x). It is easy to prove that Q(x) also has integer coeffs.

Hence for x =1,
31 = P(1)*Q(1) This means P(1)=31 or P(1) =1. P(1) = 1+b+c. Similarly for x =-1 we get similar possibilties for P(-1) = 1-b+c.

Now 3x^4+4x^2+28x+5 = P(x) Q1(x).

Hence 5 = P(0) Q1(0)
This gives c=1 or 5

Eliminating the various possibilties, we have c=1 and b =- 1 0r 1 or -31 or +31. Pls note these are just necessary conditions and themselves are not sufficient to ensure that P(x) is a divisor of the two given polynomials.

I think I will stop here. Its going nowhere. I mean we get P(1) = 31 or 1. But P(x) doesnt appear to divide the givine polynomials.

Scorching goIITian

Joined: 2 Mar 2007
Posts: 273
10 Feb 2008 23:48:28 IST
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@hsbhatt
a lil calculation mistake , for x=1, x^4 +.6x^2+25 is not 31 but 32

nd is d ans 8  ie p(1)= 8 ?

Forum Expert
Joined: 28 Feb 2007
Posts: 2185
11 Feb 2008 08:48:57 IST
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Uh Oh! Messed that one up. But, hey that method could prove useful some time!

Another attempt.

Say ,  are roots of x2+bx+c

Then  ,   are also zeros of the polynomials P and Q given in the qn. Hence they are also zeros of 3P-Q = 14x2-28x+70 = 14(x2-2x+5) and hence roots of x2-2x+5=0.

Hence x2+bx+cx2-2x+5
Hence p(x) =x2-2x+5 giving p(1) = 4.

I hope I have redeemed myself. Notice c = 5 which satisfies the condition for c in my previous post.

Forum Expert
Joined: 28 Feb 2007
Posts: 2185
12 Feb 2008 19:57:55 IST
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Hey md.... when are you going to tell us if we are right or wrong man? I am growing grey hair waiting for you to say something.

Blazing goIITian

Joined: 23 Jan 2007
Posts: 678
12 Feb 2008 22:13:05 IST
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Here is another approach to the problem:

Let and

Note that g(x) has no  term even though it is a fourth degree polynomial. Also, all the roots are complex. This hints that it can probably be factorised into quadratics.

g(x) should be as follows:

and the condition that . (since coefficient of is 0)

OR

such that.

I started checking my claims with the 2nd option (since the first one looks like it will involve a lot of calculation.)

So we have,

.

Compare co-efficients to get, , and

Hence, .

So one of the terms must be p(x). Divide and obtain.

.

Hence,      .

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