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.

His answer is correct