SOLUTION: This is from a worksheet. 6y^2+5y-4 I have to factor. I believe that the two integers I need are 8 and -3, but I don't know how to finish working out this problem.

Click here to see ALL problems on Polynomials-and-rational-expressions

Question 87645: This is from a worksheet.
6y^2+5y-4
I have to factor. I believe that the two integers I need are 8 and -3, but I don't know how to finish working out this problem.
Found 2 solutions by longjonsilver, stanbon:
Answer by longjonsilver(2297) (Show Source):
You can put this solution on YOUR website!
2 numbers that multiply to give 4 --> either 1x4 or 2x2

Added complication: the x^2 term has a 6 so either we have 6x1 or 3x2 there. We now need to mess around with these 2 pairings until we get 5.

3x2 and 1x4 could give us a 5 --> 3x1 and 2x4 could by having -3+8
no othere pairings give a 5

So (3x 4)(2x 1) is the order we need. We just need to figure out the signs. The 4 and 1 multiply to give -4 so we need 1 +ve and 1 -ve. So either:
(3x + 4)(2x - 1) or (3x - 4)(2x + 1)
Looking at the first version: -3x+8x --> +5x which is what we need, so this is the answer. The second version would give us -5x because the only difference is the order of the signs.

answer: (3x + 4)(2x - 1)

the more you do, the quicker/better you get.

cheers
Jon

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
6y^2+5y-4
I have to factor
------------------
Rewrite the problem replacing the middle term with 8y-3y:
6y^2+8y-3y-4
Factor the 1st two and the last two terms separately:
2y(3y+4)-(3y+4)
Factor again to get:
(3y+4)(2y-1)
===============
Cheers,
Stan H.