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

Ditto hardworker_indian!!

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.

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?

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..

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

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.

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

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

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.



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.

hardworker_indian, appreciate you taking time with explanation, it's been years since I've done f(x), thanks for refreshing the memory.

For what value of p is the expression 2(x*x*x)-3(x*x)+7x+p divisible by x+1 ?

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.

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?

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!

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

Good luck.

Thanks Haas...

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