Factor each polynomial completely, if a polynomial is prime say so
9x^2+4y^2
This cannot be factored without using imaginary numbers.
The answer is (3x + 2i y)(3x - 2i y)
where i is the square root of -1
if i is square root of -1
would the anwer be than
(3x+1/2y)(3x+1/2y)
To factor the polynomial 9x^2 + 4y^2, we need to determine if it can be factored further or if it is a prime polynomial.
To do this, we can check if it can be written as the product of two or more simpler expressions. In this case, the given polynomial is a sum of squares, which leads us to the identity known as the difference of squares:
a^2 - b^2 = (a + b)(a - b)
By applying this identity, we can rewrite 9x^2 + 4y^2 as (3x)^2 + (2y)^2. Now, we recognize that 9x^2 + 4y^2 can be factored further using the difference of squares:
9x^2 + 4y^2 = (3x + 2y)(3x - 2y)
Therefore, the polynomial 9x^2 + 4y^2 has been factored completely into the product of (3x + 2y) and (3x - 2y).