Answer:
(3x + 11y)^2
Step-by-step explanation:
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.
The factors of the equation are [tex]\rm (3x+11y)(3x+11y)[/tex].
Given
Equation [tex]\rm 9x^2+66xy+121y^2[/tex]
Factors of a number are numbers that divide evenly into another number.
Factorization writes a number as the product of smaller numbers.
The factors of the equation are given by using the following formula;
[tex]\rm (a+b)^2=a^2+b^2+2ab[/tex]
Then,
The factor of the equation is
[tex]\rm 9x^2+66xy+121y^2\\\\(3x)^2+(11y)^2+2(3x)(11y)\\\\(3x+11y)^2\\\\(3x+11y)(3x+11y)[/tex]
Hence, the factors of the equation are [tex]\rm (3x+11y)(3x+11y)[/tex].
To know more about factors click the link given below.
https://brainly.com/question/6810544
