Factorize $x^4 + 64$.

Given: Polynomial $x^4 + 64$.

To do: To factorize the given polynomial.

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 sqaures. Considering the pattern for this type of factorizaion $(a^2 - b^2)$, $a=x^2 + 8$ while $b = 4x$.

So, knowing that $a^2 - b^2 = (a + b)(a - b)$, we have

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