Factor x^2 - 4xy + 4y, or explain why it cannot be factored over the integers.

Step-by-step solution

1. 1

   Check for a common factor. The three terms are $x^2$, $-4xy$ and $4y$. The first has no $y$ and the last has no $x$, so there is no common variable factor, and $x^2$ has no factor of $4$. Nothing can be pulled out.

2. 2

   Test the perfect-square pattern and see it fail. The pattern $a^2-2ab+b^2$ would need the last term to be a square. But $4y$ is not a square: its exponent on $y$ is odd. The familiar identity applies to

   $x^2-4xy+4y^2=(x-2y)^2$

   and the missing exponent on the final $y$ is the whole difference.

3. 3

   Treat it as a quadratic in $x$ and compute the discriminant. Writing it as $x^2-(4y)x+4y$:

   $\Delta=(4y)^2-4(1)(4y)=16y^2-16y=16y(y-1)$

4. 4

   Draw the conclusion. A quadratic in $x$ with polynomial coefficients factors over $\mathbb{Z}[y]$ only when its discriminant is the square of a polynomial. Here $16y(y-1)$ is $16$ times a product of two distinct linear factors, which is not a perfect square in $\mathbb{Z}[y]$. Hence $x^2-4xy+4y$ is irreducible.

5. 5

   Sanity-check by partial grouping. The best that can be done is

   $x^2-4y(x-1)$

   which is a valid rewriting but not a factorisation, since the two remaining pieces share nothing. If the intended problem was $x^2-4xy+4y^2$, the answer is $(x-2y)^2$.