# What is the complete factorization of x^2-8x+16?

Jan 9, 2017

To completely factor this we need to "play" with multiples of $16$: (1x16, 2x8, 4x4, 8x2, 16x1).

This gives:

$\left(x - 4\right) \left(x - 4\right)$ or ${\left(x - 4\right)}^{2}$