Factor the expression 4x^2 + 4xy + y^2 - 4x - 2y - 3.

Step-by-step solution

1. 1

   Spot the perfect square hiding in the quadratic part. The degree-2 terms are $4x^2 + 4xy + y^2$, which matches $a^2 + 2ab + b^2$ with $a = 2x$, $b = y$:

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

   Whenever the highest-degree part of a two-variable expression is a perfect square, it is worth checking whether the rest lines up with the same block.

2. 2

   Check that the linear part is a multiple of the same block.

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

   It is — so the whole expression depends on $x$ and $y$ only through the combination $2x + y$.

3. 3

   Substitute a single variable. Let $u = 2x + y$. The expression becomes an ordinary quadratic:

   $u^2 - 2u - 3$

4. 4

   Factor the quadratic. Two numbers multiplying to $-3$ and adding to $-2$ are $-3$ and $1$:

   $u^2 - 2u - 3 = (u - 3)(u + 1)$

5. 5

   Substitute back.

   $(2x + y - 3)(2x + y + 1)$

6. 6

   Verify by expanding. $(2x+y-3)(2x+y+1) = (2x+y)^2 + (2x+y) - 3(2x+y) - 3 = (2x+y)^2 - 2(2x+y) - 3$, which is $4x^2 + 4xy + y^2 - 4x - 2y - 3$. A numerical spot check at $(x,y) = (1.7, -0.4)$ gives $-2.24$ from both forms.