Answer
$\left( x-\dfrac{1}{5}\right)^2$
Work Step by Step
The 2 numbers whose product is $ac=
1\left( \dfrac{1}{25} \right)=\dfrac{1}{25}
$ and whose sum is $b=
-\dfrac{2}{5}
$ are $\left\{
-\dfrac{1}{5}, -\dfrac{1}{5}
\right\}.$ Using these two numbers to decompose the middle term of the given trinomial, $
x^2-\dfrac{2}{5}x+\dfrac{1}{25}
,$ then the factored form is
\begin{array}{l}
x^2-\dfrac{1}{5}x-\dfrac{1}{5}x+\dfrac{1}{25}
\\\\=
\left( x^2-\dfrac{1}{5}x \right) -\left( \dfrac{1}{5}x-\dfrac{1}{25}\right)
\\\\=
x\left( x-\dfrac{1}{5} \right) -\dfrac{1}{5}\left( x-\dfrac{1}{5}\right)
\\\\=
\left( x-\dfrac{1}{5}\right)\left( x-\dfrac{1}{5}\right)
\\\\=
\left( x-\dfrac{1}{5}\right)^2
.\end{array}
