How do you factor #y^2+6y+9# using the perfect squares formula?
Aug 13, 2016
Here's how I go about it.
- Since both the middle and constant terms are positive, it could be a perfect square of the factor
#x+c#. The other option is if the middle term is negative but the constant is positive (then the factor is #x-c#).
- The middle term has a coefficient that should be twice the constant if it is to be a perfect square. (That's actually how you'd complete the square.)
- Since half of 6, squared, is 9, the constant term in the factor is half of the middle coefficient in the expanded form.
Though your variable is
Therefore, this factors into