2006-08-18 20:40:45 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
3(y^4-4y^2+4)=3(y^2-2)^2 so the factors are 3,(y^2-2),(y+(2)^1/2),(y-(2)^1/2)
2006-08-18 20:46:47 · answer #1 · answered by raj 7 · 0⤊ 0⤋
Put 3 in factor :
3(y^4-4y²+4)= 3 (y²-2)² = 3 (y²-2) (y²-2)
because (a-b)²= a²+b²-2 ab
2006-08-19 03:43:41 · answer #2 · answered by fred 055 4 · 1⤊ 0⤋
3y^4 - 12y^2 +12 = 0
3(y^4 - 4y^2 + 4) = 0
let u=y^2
3(u^2 - 4u + 4) = 0
3(u - 2)(u - 2) = 0
u = 2
therefore,
y = radical(2)
2006-08-19 03:49:40 · answer #3 · answered by PenguinMoose 3 · 0⤊ 0⤋
The factors are:
(3y^2 - 6) and (y^2 - 2)
2006-08-19 03:46:28 · answer #4 · answered by John Rae 2 · 0⤊ 0⤋
3y^4 - 12y^2 + 12
3(y^4 - 4y^2 + 4)
3(y^2 - 2)(y^2 - 2)
2006-08-19 09:46:14 · answer #5 · answered by Sherman81 6 · 0⤊ 0⤋
3 only
2006-08-19 07:58:05 · answer #6 · answered by Anonymous · 0⤊ 0⤋
if you look carefully you will see that all powers of y are even.
you can reduce the power by putting y^2 = x
now your problem is reduced to factorize an quadratic polynome in x.
For this you have the abc formula.
2006-08-19 04:18:20 · answer #7 · answered by gjmb1960 7 · 0⤊ 0⤋