[tex]\bf \begin{array}{cccccllllll}
{{ a}}^2& + &2{{ a}}{{ b}}&+&{{ b}}^2\\
\downarrow && &&\downarrow \\
{{ a}}&& &&{{ b}}\\
&\to &({{ a}} + {{ b}})^2&\leftarrow
\end{array}\qquad
% perfect square trinomial, negative middle term
\begin{array}{cccccllllll}
{{ a}}^2& - &2{{ a}}{{ b}}&+&{{ b}}^2\\
\downarrow && &&\downarrow \\
{{ a}}&& &&{{ b}}\\
&\to &({{ a}} - {{ b}})^2&\leftarrow
\end{array}[/tex]
as you can see above... notice the middle term, in a perfect square trinomial, the middle term is, 2 times the guy on the left, without the exponent, and times the guy on the right
so.. .what we have here ... let's see yours is [tex]\bf x^2-\cfrac{2}{5}x\implies x^2-\cfrac{2}{5}x+\boxed{?}^2[/tex]
so.. we have a missing guy there, the guy on the right hmm what the dickens would that be anyway?
well... let us use the middle guy [tex]\bf 2\cdot x\cdot \boxed{?}=\cfrac{2}{5}x\implies \boxed{?}=\cfrac{2x}{10x}\implies \boxed{?}=\cfrac{1}{5}[/tex]
aha ... there's our guy on the right.... so now we know is 1/5
now... let us call or very good friend Mr Zero, 0
if we "add" whatever, we also have to "subtract" whatever, since all we're really doing is, borrowing from 0
thus [tex]\bf x^2-\cfrac{2}{5}x\implies x^2-\cfrac{2}{5}x\quad +\left( \cfrac{1}{5} \right)^2\quad -\left( \cfrac{1}{5} \right)^2
\\\\\\
\left[ x^2-\cfrac{2}{5}x +\left( \cfrac{1}{5} \right)^2 \right]-\left( \cfrac{1}{5} \right)^2\implies \left( x-\cfrac{1}{5} \right)^2 - \cfrac{1^2}{5^2}
\\\\\\
\left( x-\cfrac{1}{5} \right)^2 -\cfrac{1}{25}[/tex]
