How do you 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.

##x^4 + 64 = x^4 + 16x^2 - 16x^2 + 64##

The two terms ##16x^2## and ##-16x^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 + 16x^2 - 16x^2 + 64 = x^4 + 16x^2 + 64 - 16x^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 + 16x^2 + 64 - 16x^2##

##= (x^2 + 8)(x^2 + 8) - (4x)^2##

##= (x^2 + 8)^2 - (4x)^2##

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

##(x^2 + 8)^2 - (4x)^2 = (x^2 + 8 + 4x)(x^2 + 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.