factor 9x^2+66xy+121y^2

Answer: (3x + 11y)^2

Demonstration:

The polynomial is a perfect square trinomial, because:

1) √ [9x^2] = 3x

2) √121y^2] = 11y

3) 66xy = 2 *(3x)(11y)

Then it is factored as a square binomial, being the factored expression:

 [ 3x + 11y]^2

Now you can verify working backwar, i.e expanding the parenthesis.

Remember that the expansion of a square binomial is:

- square of the first term => (3x)^2 = 9x^2
- double product of first term times second term =>2 (3x)(11y) = 66xy
- square of the second term => (11y)^2 = 121y^2

=> [3x + 11y]^2 = 9x^2 + 66xy + 121y^2, which is the original polynomial.