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by Aman verma » Tue Nov 19, 2013 1:43 am
Q: Factorize (x^4 + 4 ):

(A) (x^2 + 2)^2

(B) (x^2 + 2x + 2)(x^2 - 2x + 2 )

(C) (x^2 + 2)(x^2 - 2)

(D) (x^2 + 4)(x^2 - 4)

(E) (x^2 - 1)(x^2 + 1)


NB: Now Brent has earlier mentioned in my previous post that sum of squares can not be factored. I would like to know how much that proposition is applicable to this problem.
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by ganeshrkamath » Tue Nov 19, 2013 3:15 am
Aman verma wrote:Q: Factorize (x^4 + 4 ):

(A) (x^2 + 2)^2

(B) (x^2 + 2x + 2)(x^2 - 2x + 2 )

(C) (x^2 + 2)(x^2 - 2)

(D) (x^2 + 4)(x^2 - 4)

(E) (x^2 - 1)(x^2 + 1)


NB: Now Brent has earlier mentioned in my previous post that sum of squares can not be factored. I would like to know how much that proposition is applicable to this problem.
In the above problem, go with the options:

(A) (x^2 + 2)^2 = x^4 + 4x^2 + 4
Eliminate

(B) (x^2 + 2x + 2)(x^2 - 2x + 2 )
= (x^2 + 2 + 2x)(x^2 + 2 - 2x)
= ((x^2 + 2)^2 - (2x)^2)____________________(a+b)(a-b) = a^2 - b^2
= (x^4 + 4x^2 + 4 - 4x^2)
= (x^4 + 4)
Correct!

(C) (x^2 + 2)(x^2 - 2) = (x^4 - 4)
Eliminate

(D) (x^2 + 4)(x^2 - 4) = (x^4 - 16)
Eliminate

(E) (x^2 - 1)(x^2 + 1) = (x^4 - 1)
Eliminate

Choose B

NOTE: This problem really shows how GMAT can throw you off-guard by breaking generalized rules. Good one!

Cheers
Last edited by ganeshrkamath on Tue Nov 19, 2013 3:35 am, edited 1 time in total.
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by theCodeToGMAT » Tue Nov 19, 2013 3:17 am
Q: Factorize (x^4 + 4)

Let x = 1

(x^4 + 4) = 1^4 + 4 = 5


(A) (x^2 + 2)^2 =(3)^2 => NO

(B) (x^2 + 2x + 2)(x^2 - 2x + 2 ) =(1+2+2)(1-2+2) => 5 YES

(C) (x^2 + 2)(x^2 - 2) => (3)(-1) = -3 NO

(D) (x^2 + 4)(x^2 - 4) => (5)(-3) = -15 NO

(E) (x^2 - 1)(x^2 + 1) => (0) = NO

Answer [spoiler]{B}[/spoiler]?
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by Uva@90 » Tue Nov 19, 2013 3:19 am
Aman verma wrote:Q: Factorize (x^4 + 4 ):

(A) (x^2 + 2)^2

(B) (x^2 + 2x + 2)(x^2 - 2x + 2 )

(C) (x^2 + 2)(x^2 - 2)

(D) (x^2 + 4)(x^2 - 4)

(E) (x^2 - 1)(x^2 + 1)


NB: Now Brent has earlier mentioned in my previous post that sum of squares can not be factored. I would like to know how much that proposition is applicable to this problem.
Hi Aman Verma,
Yes, Sum of Square cannot be factored.

For this problem solve the solution given instead of solving the question.
Option A: (x^2 + 2)^2 => X^2+2+2.X^2 not as what mentioned in question.
Option B : It is shown as big, keep it aside.
Option C:(x^2 + 2)(x^2 - 2)
it is of the format
(a+b)(a-b) = (a^2-b^2)
So, (x^2 + 2)(x^2 - 2) => (X^4 - 4)Not as what mentioned in question
Option D:(x^2 + 4)(x^2 - 4)
Same as above mentioned formula
(x^2 + 4)(x^2 - 4) => (X^4-4^2),Not as what mentioned in question
Option E:(x^2 - 1)(x^2 + 1)
Same formula again
(x^2 - 1)(x^2 + 1) => (X^4-1)Not as what mentioned in question

So only Option Remaining is B

Hence Answer is B

Aside: (x^2 + 2x + 2)(x^2 - 2x + 2 ) => x^4 - 2.x.x^2 + 2x^2+ 2.x.x^2 - 4*x^2 + 4x +2x^2 - 4*x +4 => x^4+4
What is mentioned in question

Hope it helps

Regards,
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by GMATGuruNY » Tue Nov 19, 2013 3:40 am
Aman verma wrote:Q: Factorize (x^4 + 4):

(A) (x^2 + 2)^2

(B) (x^2 + 2x + 2)(x^2 - 2x + 2)

(C) (x^2 + 2)(x^2 - 2)

(D) (x^2 + 4)(x^2 - 4)

(E) (x^2 - 1)(x^2 + 1)
Let x=1.
Then x� + 4 = 1� + 4 = 5. This is our target.
Now plug x=1 into the answer choices to see which yields our target of 5.

(A) (1² + 2)² = 9. Eliminate A.

(B) (1² + 2*1 + 2)(1² - 2*1 + 2) = 5. Hold onto B.

(C) (1² + 2)(1² - 2) = -3. Eliminate C.

(D) (1² + 4)(1² - 4) = -15. Eliminate D.

(E) (1² - 1)(1² + 1) = 0. Eliminate E.

The correct answer is B.
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by Mathsbuddy » Tue Nov 19, 2013 8:13 am
I would eliminate as much as possible first time round using x = 0:

1st test x = 0: (x^4 + 4) = 4

(A) = 4 OK
(B) = 4 OK
(C) = -4 NO
(D) = -16 NO
(E) = -1 NO

Now only A and B need to be tested.

2nd test x = 1: (x^4 + 4) = 5

A is simpler, so start here
(A) (1^2 + 2)^2 = 9 NO

So the answer must be B.

Quick check:

(B) (1^2 + 2 + 2)(1^2 - 2 + 2) = 5 PERFECT.
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by Mathsbuddy » Tue Nov 19, 2013 8:19 am
You can instantly eliminate C, D and E on inspection:
The product of the free integers at the end of each bracket should equal 4.
Only A and B satisfy this condition.

Then A can be eliminated quickly because (x^2 + 2)^2 = x^4 + 4x^2 + 4 and not (x^4 + 4)

So the answer must be B.
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by Aman verma » Tue Nov 19, 2013 9:58 am
Uva@90 wrote:
Hi Aman Verma,
Yes, Sum of Square cannot be factored.
Hi Uva,

Thanks for the response. But I don't get it. We clearly have two factors for x^4 + 4 in option B, which is a sum of squares. So how come sum of squares cannot be factored. I must be missing some important concept. Please explain and elaborate. This might come handy in the test.
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by [email protected] » Tue Nov 19, 2013 1:56 pm
Hi Aman Verma,

I think that Brent was trying to point out that the GMAT won't ask you to factor the sum of squares. In your question, you didn't actually have to factor the given equation to get the answer; you could TEST Values or TEST THE ANSWERs by multiplying them together to see which one matched the given prompt.

It would be akin to use Trigonometry to solve a Geometry question; while you COULD do it, the GMAT won't require you to.

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by Uva@90 » Tue Nov 19, 2013 7:27 pm
Aman verma wrote:
Uva@90 wrote:
Hi Aman Verma,
Yes, Sum of Square cannot be factored.
Hi Uva,

Thanks for the response. But I don't get it. We clearly have two factors for x^4 + 4 in option B, which is a sum of squares. So how come sum of squares cannot be factored. I must be missing some important concept. Please explain and elaborate. This might come handy in the test.
Aman Verma,
I think Rich has mentioned a point.

Coming to the question,as far to my knowledge,I have never seen any formula for a^2 +b^2
But, as you know, we have formula for
a^2-b^2 =(a+b)(a-b)
(a+b)^2
and (a-b)^2.

In addition, still if you need to factorize,which you can do, but a tedious one.

You should add and subtract few variables,
x^4 + 4
Add and subtract below variables,
2x.x^2
4x^2(2*2x^2)
4x

so,x^4 +2x^2+2x.x^2 -2x.x^2 -4x^2 -4x +2x^2 +4x +4

Now, group similar terms,
x^2(x^2+2x+2) -2x(x^2+2x+2) +2(x^2+2x+2)

(x^2+2x+2)(x^2-2x+2)
Hence you got what we want.

I Hope you got it.

Dude you made me to work a lot :P

Regards,
Uva.
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by Mathsbuddy » Wed Nov 20, 2013 12:03 am
Some people may just notice that B is correct, as per the reasoning below.
However, I wouldn't recommend trying to find solutions this way, when the other methods could be quicker in the end.

Let p = x^2 + 2 and q = 2x


Then (B) (x^2 + 2x + 2)(x^2 - 2x + 2) = (p + q) * (p - q)

which is the difference of 2 squares = (p^2 - q^2) = (x^2 + 2)^2 - (2x)^2 = (x^4 + 4)
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by Aman verma » Wed Nov 20, 2013 1:30 am
EDITED with correct response:

Hi all !

Many thanks for the responses. Rich has made no point at all. Brent never mentioned that GMAT will not ask to factor sum of squares. In fact Brent has removed the post he made by editing . He himself mentioned that there was an error. That's the mark of a real GMAT expert. I really appreciate him for doing that. So I don't know how Rich cook up that story.

Coming to Uva I don't know what he is trying to prove by self contradicting himself. At one place he is claiming that sum of squares cannot be factored and at the very next stance he is himself factoring sum of squares. This is so self contradicting and self defeating. I don't know what he is getting at by self contradicting himself. This is so weird and self defeating. Also why he factored the expression when other students have already shown the factors in option (B). The entire exercise by Uva was redundant and self defeating. He was unable to prove that sum of squares cannot be factored. I still confirm my stand that it is an over-generalisation to claim that sum of squares cannot be factored. Which was confirmed by Brent by the edit he made.

Regards
Last edited by Aman verma on Thu Nov 28, 2013 3:37 am, edited 2 times in total.
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by Gmat Bond » Thu Nov 28, 2013 1:54 am
Aman verma wrote:Hi all !

Many thanks for the responses. Rich has made a very good point that the GMAT will not require to factor a sum of squares though sum of squares can be factored. I think it was an overgeneralisation to claim that sum of squares can not be factored. Nevertheless we don't require to bother about it for the GMAT as it will not ask us to factor sum of squares. Thank you all for clarifying the concept and I will not automatically reject an equation(sum of squares) from factorization. I look forward for continuing guidance of you all.

Thanks
Hey What's the fuss! Why you guys are repeating the same thing again an again. You are repeating your own words. Uva is confirming what you are saying. What's going on? Are you guys playing some repeatition game ? What's going on ?
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by Aman verma » Thu Nov 28, 2013 3:36 am
Gmat Bond wrote:
Hey What's the fuss! Why you guys are repeating the same thing again an again. You are repeating your own words. Uva is confirming what you are saying. What's going on? Are you guys playing some repeatition game ? What's going on ?
Hi Gmat Bond,

I am not playing any repeatition game. I was only reiterating my point. As far as others are concerned I cannot say. In fact I have made an EDIT in my post which does not repeat anything and clarifies my position. Check the edit!
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