+0  
 
0
62
1
avatar

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
avatar
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

14 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.