A little bit of trained eyeballing can help here. Let me tell you how I saw the answer without any computation.
Remember that the expansion of a squared binomial is
[tex] (a+b)^2 = a^2+2ab+b^2 [/tex]
So, the square of a binomial has the following properties, which you can easily spot:
- It has three terms
- Two of them are perfect squares
- The third is twice the product of the two roots.
Well, in this case we do have three terms, and indeed [tex] 121x^2 [/tex] is the square of 11x, and 1 is the square of itself. The only thing we have to check is that 22x is twice the product of 11x and 1, which is true.
So, the answer is
[tex] 121x^2-22x+1 = (11x-1)^2 [/tex]
If you want to see some explicit calculations, just use the quadratic formula
[tex] x_{1,2} = \cfrac{-b\pm\sqrt{b^2-4ac}}{2a} [/tex] to see that there is a double solution [tex] x = \frac{1}{11} [/tex], and then use the formula
[tex] ax^2+bx+c = a(x-x_1)(x-x_2) [/tex] to come to the same conclusion.
