Answer
$x^2-8x+16=(x-4)^2$
Work Step by Step
We add the square of half of the coefficent of $x$ so that the result is a perfect square trinomial.
Co-efficient of $x=-8$
Half of -8 is $\frac{1}{2}×-8=-4$
Square of -4 is $-4×-4=16$
We add 16 to $x^2-8x$ to make it a perfect square trinomial.
Hence it becomes $x^2-8x+16$
Factored form-
$x^2-8x+16$
$=x^2-4x-4x+16$
$=x(x-4)-4(x-4)$ (Taking the common factors)
$=(x-4)(x-4)$ ($ x-4$ is taken common from both the terms)
$=(x-4)^2$
