Rewrite (2x + 1)² + 2(2x + 1) + 1 as a perfect square.

Step-by-step solution

1. 1

   Look for the shape before touching the algebra. The three terms are: something squared, twice that same something, and $1$. That is exactly the perfect-square pattern

   $A^2 + 2A + 1 = (A+1)^2$

   Recognising it saves expanding to $4x^2 + 8x + 4$ and re-factoring afterwards.

2. 2

   Name the repeated block. Set

   $A = 2x + 1$

   Then the expression is literally $A^2 + 2A + 1$. The middle term must be $2A$ (not $2$ times something else) for the pattern to apply — here $2(2x+1) = 2A$, so it does.

3. 3

   Apply the identity and substitute back.

   $A^2 + 2A + 1 = (A+1)^2 = \bigl((2x+1) + 1\bigr)^2 = (2x+2)^2$

4. 4

   Pull the common factor out of the bracket. Both terms inside share a factor $2$:

   $(2x+2)^2 = \bigl(2(x+1)\bigr)^2 = 4(x+1)^2$

   The squared form $4(x+1)^2$ is usually the preferred final answer because the bracket is fully reduced.

5. 5

   Verify by expanding both sides. The original expands to

   $(4x^2 + 4x + 1) + (4x + 2) + 1 = 4x^2 + 8x + 4$

   and the answer expands to $4(x^2 + 2x + 1) = 4x^2 + 8x + 4$. They agree, and a spot check at $x = 3$ gives $49 + 14 + 1 = 64 = 4(4)^2$.