I just want to check if what I did is right, so here's the trinomial: \({2y^2 - 5y - 3}\)

Here's what I did:

\({2y^2 - 6y + y - 3}\)

\({2y (y - 3)(y - 3)}\) (factor out (y - 3))

\({2y (y-3)}\)

Guest Apr 2, 2018

#1**0 **

Not quite:

Factor the following:

2 y^2 - 5 y - 3

Factor the quadratic 2 y^2 - 5 y - 3. The coefficient of y^2 is 2 and the constant term is -3. The product of 2 and -3 is -6. The factors of -6 which sum to -5 are 1 and -6. So 2 y^2 - 5 y - 3 = 2 y^2 - 6 y + y - 3 = y (2 y + 1) - 3 (2 y + 1):

y (2 y + 1) - 3 (2 y + 1)

Factor 2 y + 1 from y (2 y + 1) - 3 (2 y + 1):

**=(2y + 1) (y - 3)**

Guest Apr 2, 2018