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Cool goIITian

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10 Feb 2008 13:31:17 IST

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648

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

Comments (7)

Rahul Raghavendra

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10 Feb 2008 13:34:23 IST

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what is the last eqn

Rahul Raghavendra

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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

Hari Shankar

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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

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.

shine

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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 ?

Hari Shankar

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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+c

x²-2x+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.

Hari Shankar

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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.

Abhijith

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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)

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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