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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))

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

 Apr 2, 2018
 #1
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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)

 Apr 2, 2018

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