Factor 9x^2 + 6x + 1.

Step-by-step solution

1. 1

   Check whether the outer terms are perfect squares. They are:

   $9x^2=(3x)^2,\qquad 1=1^2$

   so the candidate pattern is $a^2+2ab+b^2=(a+b)^2$ with $a=3x$ and $b=1$.

2. 2

   Verify the middle term — this is the step that decides it. The pattern demands $2ab$:

   $2ab=2(3x)(1)=6x$

   which matches the given middle term exactly. Without this check you could wrongly 'factor' something like $9x^2+7x+1$, which is not a perfect square at all.

3. 3

   Write the factorisation.

   $9x^2+6x+1=(3x+1)^2$

4. 4

   Confirm by expanding back.

   $(3x+1)^2=9x^2+3x+3x+1=9x^2+6x+1\qquad\checkmark$

   A second confirmation: the discriminant is $6^2-4(9)(1)=0$, and a zero discriminant is exactly the algebraic signature of a perfect square.