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# For what value of p is the expression 2(x*x*x)-3(x*x)+7x+p

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VP
Joined: 25 Nov 2004
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For what value of p is the expression 2(x*x*x)-3(x*x)+7x+p [#permalink]  12 Jun 2006, 13:17
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For what value of p is the expression 2(x*x*x)-3(x*x)+7x+p divisible by x+1 ?
Manager
Joined: 10 May 2006
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Location: USA
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Kudos [?]: 3 [0], given: 0

P = 12

I basically took 2(x^3)-3(x^2)+7x+p and divded it by x+1 (long division style).

I ended up with 2(x^3)-3(x^2)+7x+12 = [2(x^2) -5x +12] * [x+1]

P needs to be 12 in order to divide x+1 into the equation evenly.
VP
Joined: 02 Jun 2006
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Kudos [?]: 48 [0], given: 0

This one was a weird one for me... didn't expect cubic equation division to come in GMAT.

Well the answer (long way again... as tl mentioned) is p= 12.

The bigger question I have is should I be prepared to expect such a question on the GMAT?
Manager
Joined: 10 May 2006
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In my opinion, because this question tested creative thinking more so then any specific properties of the cubic square, I think that it's fair game. I had to think for a little while before realizing that straight division would work.

The first time I took the gmats, I didn't remember seeing anything like this.....who knows though...I'll have a better gauge this Sat after I attempt the Gmats for the second time!
VP
Joined: 02 Jun 2006
Posts: 1266
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Kudos [?]: 48 [0], given: 0

Given that you solved this, I am sure you are more than prepared for anything they throw at you.

Good luck.

tl372 wrote:
In my opinion, because this question tested creative thinking more so then any specific properties of the cubic square, I think that it's fair game. I had to think for a little while before realizing that straight division would work.

The first time I took the gmats, I didn't remember seeing anything like this.....who knows though...I'll have a better gauge this Sat after I attempt the Gmats for the second time!
Manager
Joined: 10 May 2006
Posts: 186
Location: USA
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Kudos [?]: 3 [0], given: 0

Thanks Haas...
Senior Manager
Joined: 11 May 2006
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for 2(x*x*x)-3(x*x)+7x+p to be divisible by x+1,

x+1 should be a root of 2(x*x*x)-3(x*x)+7x+p = 0

subsitute x=-1 and you get p = 12
SVP
Joined: 30 Mar 2006
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Long divison style

__2x^2-5x+12_____________
x +1 )2x^3 - 3x^3 + 7x + P
Manager
Joined: 11 Oct 2005
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I prefer the root method - easy, quick and painless - yields the same answer p = 12
Senior Manager
Joined: 16 Apr 2006
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Quote:
for 2(x*x*x)-3(x*x)+7x+p to be divisible by x+1,

x+1 should be a root of 2(x*x*x)-3(x*x)+7x+p = 0

subsitute x=-1 and you get p = 12

This solution is fast.
GMAT Club Legend
Joined: 07 Jul 2004
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2(x*x*x)-3(x*x)+7x+p = 2x^3 - 3x^2 + 7x + p

p must be 12 for the expression to be divisible by (x+1) --> Do a standard division to find this out.
VP
Joined: 29 Dec 2005
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Re: PS: easy one [#permalink]  13 Jun 2006, 18:46
MA wrote:
For what value of p is the expression 2(x*x*x)-3(x*x)+7x+p divisible by x+1 ?

i prefer to do this way: 2(x*x*x)-3(x*x)+7x+p divisible by x+1

= 2x^3 - 3x^2 + 7x + p
= 2x^3 + 2x^2 - 5x^2 - 5x + 12x + 12

(start from the begaining ==> adjust the second term accordingly so that 2x^2 is a common term from the first and second term and also make necessary adjustments accordingly so the x+1 is left from every two terms after taking a common term as above)

= 2x^2 (x+1) - 5x (x+1) + 12 (x+1)
= (x +1) (2x^2 - 5x + 12)

so P= 12.
Re: PS: easy one   [#permalink] 13 Jun 2006, 18:46
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