#### Answer

$|x-4|$

#### Work Step by Step

Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\sqrt{x^2-8x+16}
\\\\=
\sqrt{(x-4)^2}
.\end{array}
Using $\sqrt[n]{x^n}=|x|$ if $n$ is even and $\sqrt[n]{x^n}=x$ if $n$ is odd, then
\begin{array}{l}\require{cancel}
\sqrt{(x-4)^2}
\\\\=
|x-4|
.\end{array}