Please help!
2007-08-27 20:39:17 · 7 answers · asked by justme 4 in Science & Mathematics ➔ Mathematics
(2 y ²) (y^4 - 16)
(2 y ²) (y ² - 4) (y ² + 4)
(2 y ²) (y - 2) (y + 2) (y ² + 4)
2007-08-27 21:43:51 · answer #1 · answered by Como 7 · 2⤊ 0⤋
Factor out a 2y^2 which will result in
2y^2 ( y^4 - 16)
Then factor the quantity. . .
2y^2 (y^2 - 4) ( y^2 + 4)
Yah. . .that's it
2007-08-28 03:44:25 · answer #2 · answered by Anonymous · 0⤊ 1⤋
2y^6 - 32y^2 = 2y^2(y^4 - 16)
=2y^2(y^2-4)(y^2+4)
=2y^2(y-2)(y+2)(y^2+4)
is this correct?
2007-08-28 09:27:22 · answer #3 · answered by karl-a 2 · 0⤊ 1⤋
Factoring: 2y^6 - 32y^2
---------------------------
2y^6 - 32y^2
= 2y^2(y^4 - 16)
= 2y^2(y^2-4)(y^2+4)
= 2y^2(y-2)(y+2)(y-j2)(y+j2)
where: j = square root of -1
Or just leave as:
= 2y^2(y^2+4)(y-2)(y+2)
2007-08-28 04:00:57 · answer #4 · answered by taq2007 2 · 0⤊ 1⤋
= 2y^2 (y^4-16)
= 2y^2 [(y^2)^2-4^2)
= 2y^2 (y^2-4) (y^2+4)
2007-08-28 03:57:16 · answer #5 · answered by isya c 1 · 0⤊ 1⤋
2y² (y^4 - 16)
2y² (y² + 4) (y² - 4)
2007-08-28 03:59:28 · answer #6 · answered by hznfrst 6 · 0⤊ 1⤋
2y^2(y^4-16)
2y^2(y^2+4)(y^2-4)
2y^2(y^2+4)(y+2)(y-2)
2007-08-28 03:50:31 · answer #7 · answered by mrpscience 1 · 0⤊ 0⤋