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# If (x^2+5x-1)^2-25=(x-a)(x-b)(x-c)(x-d) and a,b,c, and d are

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Intern
Joined: 05 Apr 2008
Posts: 23
If (x^2+5x-1)^2-25=(x-a)(x-b)(x-c)(x-d) and a,b,c, and d are [#permalink]

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13 Sep 2008, 16:46
If (x^2+5x-1)^2-25=(x-a)(x-b)(x-c)(x-d) and a,b,c, and d are constants, what is the value of abcd?

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Manager
Joined: 09 Jul 2007
Posts: 235

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13 Sep 2008, 16:53

(x^2+5x-1)^2-25=(x-a)(x-b)(x-c)(x-d)

the only non-x terms in the left hand side of expression are 1 ( as it is squared ) and -25 , so the non-x term will be -24
the right hand side of the expression the non-x term will be adcd

Intern
Joined: 05 Apr 2008
Posts: 23

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13 Sep 2008, 17:00
I still don't see it. How did you get to that solution?
Manager
Joined: 09 Jul 2007
Posts: 235

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13 Sep 2008, 17:09
If you expand (x^2+5x-1)^2 => then the expression will have terms with x^4, X^3, X^2 , X and only one non-x term.. which will be 1^2=1

so, the left hand of the expression has some terms with x^4, X^3, X^2 , X and one constant 1-25=-24
similarly the right hand of the expression will have one non-x term abcd..

considering that each constant terms will be equal , adcd=-24

you might argue that what if the right and left side of the expression are not exactly similar.. for eg x^2-1x+k=X^2+5x-j
which is possible, but in that case we can solve this problem..
VP
Joined: 17 Jun 2008
Posts: 1477

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15 Sep 2008, 00:37
ssandeepan wrote:

(x^2+5x-1)^2-25=(x-a)(x-b)(x-c)(x-d)

the only non-x terms in the left hand side of expression are 1 ( as it is squared ) and -25 , so the non-x term will be -24
the right hand side of the expression the non-x term will be adcd

Nice approach.

I approached using a^2 - b^2 factors.

The expression in the question will be
(x^2 + 5x -1 + 5)(x^2 + 5x -1 - 5)
=(x^2+5x+4)(x^2+5x-6)
=(x+1)(x+4)(x+3)(x-2)

Hence, abcd = -24.
Intern
Joined: 02 Sep 2008
Posts: 45

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15 Sep 2008, 07:59
I also followed the same approach.. X^2 - y^2.

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Re: Math Question   [#permalink] 15 Sep 2008, 07:59
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# If (x^2+5x-1)^2-25=(x-a)(x-b)(x-c)(x-d) and a,b,c, and d are

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