Ricardo O. answered • 08/30/12
service for help in Algebra, Geometry, Pre-Calculus and Calculus
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 (a2 - b2), a equals x2 + 8 while b = 4x. So, knowing that a2 - b2 = (a + b)(a - b), 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.
Roman C.
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:
(a^2+b^2)(c^2+d^2)=(ac)^2+(ad)^2+(bc)^2+(bd)^2=(ac)^2+2abcd+(bd)^2+(ad)^2-2abcd+(bc)^2=(ac+bd)^2+(ad-bc)^2
09/27/12