I'd suggest you get rid of the fraction first. Mult. every term by 3. You will get
3y^2 + 10y + 3 = 0
From inspection, with coefficient a=3 and coeff. c = 3, the binomial factors could possibly begin with y or 3y: for example, y+1; also, the binom. factors may end in +1. Let's try the possible binomial factor 3y + 1.
Note that 10y separates into 9y+1y.
Then 3y^2 + 10y + 3 = 0 becomes
3y^2 + 9y + 1y + 3 = 0
Let's apply factoring by grouping:
3y^2 + 9y + 1y + 3 = 0
3y*(y + 3) + 1(y + 3) so y+3 is indeed a common factor.
Factoring y+3 out, we get (y+3)(3y + 1), which prove to be the correct set of factors. Multiply these together to ensure that the product is indeed y^2 + (10/3)y + 1.
