How do you factor completely $25{x^2} - 10x + 1$?

Hint: We will first use the method of splitting the middle term and then, we will just take 5x common from the first two terms and then take out – 1 common from last 2 terms.

Complete step-by-step answer:
We are given that we need to completely factor the given quadratic equation $25{x^2} - 10x + 1$.
We can write this equation as following expression:-
$ \Rightarrow 25{x^2} - 5x - 5x + 1$
We can also write this equation as follows:-
$ \Rightarrow \left( {25{x^2} - 5x} \right) + \left( { - 5x + 1} \right)$
We will now take 5x common from first two terms in the first bracket, then we will obtain the following equation:-
$ \Rightarrow 5x\left( {5x - 1} \right) + \left( { - 5x + 1} \right)$
Now, we will just take – 1 common out of two terms in the second bracket, then we will obtain the following equation:-
$ \Rightarrow 5x\left( {5x - 1} \right) - 1\left( {5x - 1} \right)$
Now, we have (5x – 1) common from both the terms, then we will get the following equation:-
$ \Rightarrow \left( {5x - 1} \right)\left( {5x - 1} \right)$
We can write this as follows:-
$ \Rightarrow {\left( {5x - 1} \right)^2}$

Thus, we have the required factors.

Note:
The students must note that there is an alternate way to find the answer to the same question.
Alternate way 1:
We are given that we need to completely factor the given quadratic equation $25{x^2} - 10x + 1$.
We can also write this equation as: ${\left( {5x} \right)^2} - 2 \times 5x \times 1 + {1^2}$. ………….(1)
Now, we know that we have an identity given by the following formula:-
$ \Rightarrow {(a - b)^2} = {a^2} + {b^2} - 2ab$
Replacing a by 5x and b by 1, we will then obtain the following equation:-
$ \Rightarrow {\left( {5x - 1} \right)^2} = {\left( {5x} \right)^2} - 2 \times 5x \times 1 + {1^2}$
Putting this in equation number 1, we will then obtain the following equation:-
We can also write this equation as: ${\left( {5x - 1} \right)^2}$
Thus we have the required factors.
Alternate way 2:
We can also use the quadratic formula to find the roots.
The equation $a{x^2} + bx + c = 0$ has roots given by: $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Comparing the given equation, we have a = 25, b = - 10 and c = 1
$ \Rightarrow x = \dfrac{{10 \pm \sqrt {{{10}^2} - 4 \times 25} }}{{2 \times 25}}$
Simplifying the calculations, we will then obtain:-
$ \Rightarrow x = \dfrac{{10 \pm \sqrt {100 - 100} }}{{50}}$
Simplifying the calculations further, we will then obtain:-
$ \Rightarrow x = \dfrac{1}{5}$
Thus, we have the factors as $\left( {x - \dfrac{1}{5}} \right)$ twice.
Thus, we get ${\left( {x - \dfrac{1}{5}} \right)^2} \equiv {\left( {5x - 1} \right)^2}$.