How to factor this?

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.

*x*^{4} + 64 = *x*^{4} + 16*x*^{2} - 16*x*^{2} + 64

The two terms 16*x*^{2} and ^{-}16*x*^{2} appear by considering the square root of
*x*^{4} 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:

*x*^{4} + 16*x*^{2} - 16*x*^{2} + 64 = *x*^{4} + 16*x*^{2} + 64 - 16*x*^{2}

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.

*x*^{4} + 16*x*^{2} + 64 - 16*x*^{2}

= (*x*^{2} + 8)(*x*^{2} + 8) - (4*x*)^{2}

= (*x*^{2} + 8)^{2} - (4*x*)^{2}

The preceding expression is a difference of two sqaures. Considering the pattern for this type of factorizaion (*a*^{2} - *b*^{2}), *a* equals *x*^{2} + 8 while *b* = 4*x*. So, knowing that *a*^{2} - *b*^{2} = (*a* + *b*)(*a* - *b*), we have

(*x*^{2} + 8)^{2} - (4*x*)^{2} = (*x*^{2} + 8 + 4*x*)(*x*^{2} + 8 - 4*x*),

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.

## Comments

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