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# factor x^4+64

Most people trying to factor this expression would say that it is irreducible, that is, it cannot factor at all. I will show you a trick that consist of including two terms that not appear in the original expression.

x4 + 64 = x4 + 16x2 - 16x2 + 64

The two terms 16x2 and -16x2 appear by considering the square root of x4 and by dividing the original constant by the leading exponent (64 ÷ 4 = 16). Then, you change the position of the negative term to the last position:

x4 + 16x2 - 16x2 + 64 = x4 + 16x2 + 64 - 16x2

The first three terms form a perfect square trinomial, which can be easily factored. The last term can be rewritten considering the square root concept.

x4 + 16x2 + 64 - 16x2

= (x2 + 8)(x2 + 8) - (4x)2

= (x2 + 8)2 - (4x)2

The preceding expression is a difference of two sqaures. Considering the pattern for this type of factorizaion (a2b2), a equals x2 + 8 while b = 4x. So, knowing that a2b2 = (ab)(ab), we have

(x2 + 8)2 - (4x)2 = (x2 + 8 + 4x)(x2 + 8 - 4x),

which gives the solution to the exercise.

I'll hope that this helping tool would lead you to solve any similar type of factorization exercises.

The trick of adding and subtracting a term that doesn't appear is actually commonplace. For the factoring, I was asked for all integers x such that x^4+4 was prime. The solution was to factor it in the same way as the above: x^4+4=x^4+4x^2+4-4x^2=(x^2+2)^2-(2x)^2=(x^2+2x+2)(x^2-2x+2)=[(x+1)^2+1][(x-1)^2+1]. Clearly, both factors are >1 if x=0 or |x|>1 so the answer is just 1 and -1.

Another example is proving that sums of two squares are closed under multiplication:

9/27/2012

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