Factor completely a^2y^2-1?

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2006-12-06 01:48:43 · 5 answers · asked by ME 3 in Science & Mathematics ➔ Mathematics

This is just like the other one you did, (ay-1)*(ay+1)

2006-12-06 01:50:35 · answer #1 · answered by moto 3 · 0⤊ 0⤋

a^2.y^2 - 1 = (ay)^2 - 1 = (ay+1)(ay-1)

2006-12-06 10:07:12 · answer #2 · answered by yasiru89 6 · 0⤊ 0⤋

aÂ²yÂ² - 1

(ay + 1)(ay - 1)

- - - - -s-

2006-12-06 10:32:49 · answer #3 · answered by SAMUEL D 7 · 0⤊ 0⤋

(a^y-1)(a^y+1)(a^2y+1)

2006-12-06 10:01:14 · answer #4 · answered by ash v 3 · 0⤊ 0⤋

(ay+1)*(ay-1)

2006-12-06 09:55:29 · answer #5 · answered by azeem 2 · 0⤊ 0⤋