Factor a^2 + 4a + 4.

Step-by-step solution

1. 1

   Identify the two squares. The leading term is already a square and so is the constant:

   $a^2=(a)^2,\qquad 4=2^2$

2. 2

   Match the middle term against $2ab$. With $a$ and $b=2$:

   $2\cdot a\cdot 2=4a$

   which is exactly what appears, so the trinomial fits $a^2+2ab+b^2$.

3. 3

   Apply the identity.

   $a^2+4a+4=(a+2)^2$

4. 4

   Cross-check with ordinary factoring. Looking for two numbers with product $4$ and sum $4$ gives $2$ and $2$, so the factorisation is $(a+2)(a+2)$ — the same answer, which is why a repeated factor pair always signals a perfect square.

5. 5

   Expand to confirm. $(a+2)^2=a^2+2a+2a+4=a^2+4a+4$ $\checkmark$. Substituting a value is another quick test: at $a=3$ the trinomial gives $9+12+4=25$ and $(3+2)^2=25$ $\checkmark$.