factor completely. 2y^ - 8?

2 ⤊

2007-05-02 17:06:29 · 4 answers · asked by deborah n 1 in Science & Mathematics ➔ Mathematics

Is this 2y^2 -8
If it is then the factoring is
2(y^2-4)
2(y-2)(y+2)

2007-05-02 17:10:28 · answer #1 · answered by lizzie 3 · 1⤊ 0⤋

Assume 2yÂ² - 8:-
2.(yÂ² - 4) = 2.(y - 2).(y + 2)

2007-05-03 07:55:13 · answer #2 · answered by Como 7 · 0⤊ 0⤋

2y^ - 8
= 2/(y^8)

or did u mean 2y^2 - 8?
If yes, then
= 2(y^2 - 4)
= 2(y^2 - 2^2)
= 2(y+2)(y-2)

2007-05-03 00:12:38 · answer #3 · answered by QiQi 3 · 0⤊ 0⤋

What's the number after the ^?
As written, you get 2(y^ -4).

2007-05-03 00:10:42 · answer #4 · answered by steiner1745 7 · 1⤊ 0⤋