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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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For what value of p is the expression 2(x*x*x)-3(x*x)+7x+p [#permalink] New post 25 Aug 2004, 13:51
For what value of p is the expression 2(x*x*x)-3(x*x)+7x+p divisible by x+1 ?
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 [#permalink] New post 25 Aug 2004, 17:39
Yes, its correct. Please post your steps if possible. Thanks.
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 [#permalink] New post 25 Aug 2004, 19:12
smcgrath12 wrote:
Yes, its correct. Please post your steps if possible. Thanks.


For any polynomial experssion f(x), if x-a is a factor, then f(a)=0

So, f(x) = 2x^3 - 3x^2+7x+p
f(-1) = -2 - 3 -7 + p = 0
p = 12
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 [#permalink] New post 26 Aug 2004, 05:46
Ditto hardworker_indian!!
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 [#permalink] New post 26 Aug 2004, 05:49
you can also just do straight division:

(2x^3 - 3x^2 + 7x + p)/(x+1)

I can't do it here, but if you use the regular division bar (the one that looks like a square root) you can solve this and get 2x^2 - 5x + 12. When you do, the only number for p that will cancel out perfectly and make it work is 12.
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 [#permalink] New post 26 Aug 2004, 06:24
Some of you are just awesome. I was betting that nobody could answer that, because it goes to the Remainder Theorem and synthetic division concept, which many people won't remember from their algebra days. Is this concept explained in any standard GMAT guide?
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 [#permalink] New post 26 Aug 2004, 08:06
hi .. could u pls tell me why does not 1 work here..if you substitute x=1, even then the number is divisible by x+1.
pls explain..
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 [#permalink] New post 26 Aug 2004, 08:18
hardworker_indian wrote:
smcgrath12 wrote:
Yes, its correct. Please post your steps if possible. Thanks.


For any polynomial experssion f(x), if x-a is a factor, then f(a)=0

So, f(x) = 2x^3 - 3x^2+7x+p
f(-1) = -2 - 3 -7 + p = 0
p = 12



hardworker_indian,

Does this rule hold true for (x+a) as well? Since you are taking a=-1, shouldn't we thinking as "if (x+a) factor of ...." ?

Thanks
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 [#permalink] New post 26 Aug 2004, 10:21
Yes, for any function f(x), x+a is a factor if f(-a)=0 or x-a is a factor if f(a)=0.
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 [#permalink] New post 26 Aug 2004, 18:47
hardworker_indian wrote:
smcgrath12 wrote:
Yes, its correct. Please post your steps if possible. Thanks.


For any polynomial experssion f(x), if x-a is a factor, then f(a)=0

apparently my algebraic skills aren't where they should be... I couldn't follow this explanation
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 [#permalink] New post 26 Aug 2004, 21:32
2x^3 - 3x^2 + 7x +p will equate to 0 if x+1 is a root.

So 2(-1)^3 - 3(-1)^2 +7(-1) + p = 0
p = 12
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 [#permalink] New post 27 Aug 2004, 07:17
neelam wrote:
hi .. could u pls tell me why does not 1 work here..if you substitute x=1, even then the number is divisible by x+1.
pls explain..

If you substitute x=1,
=> x=1
=> x-1=0
we are not looking to see if x-1 is a factor. We are looking to see if x+1 is a factor. As a simple rule, you can equate the factor (x+1) to zero (x+1=0) and find the value of x (x+1=0 => x=-1) and then subsitute this in the original expression.


lastochka wrote:
apparently my algebraic skills aren't where they should be... I couldn't follow this explanation

I will try to explain as much as I can - from little notes and distant memory.

We generally note algeraic expressions as functions, f(x) -- this means that f(x) is a polynomial in x and for each value of x, f(x) gives you a result.

Rules:
For any polynomial f(x):
1. To find factors, solve: f(x)=0
2. For any value a, f(a) gives the reminder when f(x) is divided by x-a.

#2 inutrn means,
For any factor x-a, f(a) gives 0
For any nonfactor x-a, f(a) gives reminder


Lets take an example: f(x) = x^2-5x+6

To find factors:
equate f(x) = 0,
x^2-5x+6=0
(x-3)(x-2)=0
x=3 or 2

To see if x-3 is a factor:
x-3=0 => x=3
f(x) = x^2-5x+6
f(3) = 3^2-5.3+6
f(3) = 0
So, x-2 is a factor of x^2-5x+6

To see if x+5 is a factor:
x+5=0 => x=-5
f(x) = x^2-5x+6
f(5) = 5^2+5.5+6
f(3) = 56
So, x+5 is NOT a factor of x^2-5x+6.
And when you divide x^2-5x+6 by x+5, you will get a reminder of 56.

I think this should be enough for this topic, but algebra in itself is a ocean. :-D
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 [#permalink] New post 27 Aug 2004, 08:49
hardworker_indian, appreciate you taking time with explanation, it's been years since I've done f(x), thanks for refreshing the memory.
  [#permalink] 27 Aug 2004, 08:49
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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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