Factors

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.

Brent is right if the situation is a^2 + b^2 and 2ab is not a perfect square. But here the case is all different. Our 2ab is a perfect square; hence we can go for it.

Plugging values is the best approach when stuck for a missing concept. That has already been shown by few.

For your curiosity...

x^4 + 4

= x^4 + 4x^2 + 4 – 4x^2

= (x^2 + 2)^2 – (2x)^2 {this changed to a^2 - b^2 form}

= (x^2 + 2 – 2x)( x^2 + 2 + 2x)

You wouldn't see this question on the GMAT (at least at present) because it involves a variation on a technique known as 'completing the square'. Here, if you were used to completing the square, you would notice that

x� + 4

is really

(x² + 2)² - 4x²

Now you use the difference of squares to find

(x² + 2 + 2x) * (x² + 2 - 2x)

which can be rearranged into answer B.

Again, this is NOT a GMAT question: it's trivially easy if you know how to complete the square and annoying/impractical if you don't. The GMAC doesn't seem to like questions like that (nor should they).

To answer your other question, you're correct that there isn't a nice general factorization of the sum of squares -- the only general one involves complex numbers, which aren't on the GMAT -- but this specific polynomial just happens to factor nicely.