2006-10-08 08:58:51 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
3(2y-1) (y+1)
2006-10-08 09:01:18 · answer #1 · answered by Anonymous · 1⤊ 0⤋
Factor out a 3: 3(2y^2+y-1).
You could use the quadratic formula to obtain the roots of this polynomial and write the factorization that way, or you can try to factor this directly. Because of the 2y^2, you know that such a factorization would have the form
3(2y+a)(y+b),
where a+2b=1, and ab=-1. Since ab is negative, a and b have opposite signs. Try b=1 and a=-1, and that works. So you have
3(2y-1)(y+1).
2006-10-08 16:01:53 · answer #2 · answered by James L 5 · 0⤊ 0⤋
6y^2 + 3y-3 = 0
(6y -3 ) ( y+1) = 0
y = 1/2
y = -1
2006-10-08 16:11:12 · answer #3 · answered by visal m 1 · 0⤊ 0⤋
6y^2+3y-3=3(2y^2+y-1)
3(2Y-1)(y+1)
2006-10-08 16:07:29 · answer #4 · answered by yupchagee 7 · 0⤊ 0⤋
find the roots in the equation
0=6^2+3y-3
and use the result!
2006-10-08 16:01:39 · answer #5 · answered by Broden 4 · 0⤊ 0⤋
y(6y + 3-3)
y(6y + 0)
2006-10-08 16:01:02 · answer #6 · answered by LaCosaMasBella 3 · 0⤊ 0⤋
6 ( y - 1/2 ) ( y + 1 )
factor out the six and then F.O.I.L.
2006-10-08 16:01:31 · answer #7 · answered by stylesofbeyond40 1 · 0⤊ 0⤋
use the formula
a=6
b=3
c=-3
you get
y= -2.25 or -3.75
2006-10-08 16:04:06 · answer #8 · answered by idc_bd k 2 · 0⤊ 0⤋