Factor completely y^3-12y^2+36y
2006-10-17 01:21:21 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
y(y^2 - 12y + 36)
y(y - 6)(y - 6)
2006-10-17 01:29:36 · answer #1 · answered by Bird 2 · 0⤊ 0⤋
First factor out a y:
y*(y^2-12y+36)
Then factor the rest:
y*((y-6)^2)
2006-10-17 01:31:14 · answer #2 · answered by ? 4 · 1⤊ 1⤋
y(y^2-12y+36)
factoringy^2-12y+36
(y-6)^2
so the answer is
y(y-6)^2
2006-10-17 01:37:45 · answer #3 · answered by raj 7 · 1⤊ 1⤋
y³ -12y² + 36y
y(y² - 12y + 36
y( y - 6)( y - 6)
-- - - - - - - - - - - - -s
2006-10-17 03:46:50 · answer #4 · answered by SAMUEL D 7 · 0⤊ 0⤋
y=0,-6,+6
2006-10-17 01:28:07 · answer #5 · answered by oee22 2 · 0⤊ 3⤋
y(y - 6)(y - 6)
oh...its already solve..hehehe
2006-10-17 01:34:41 · answer #6 · answered by Beans 2 · 0⤊ 0⤋
(Y)(Y-6)(Y-6)
2006-10-17 01:31:30 · answer #7 · answered by Anonymous · 0⤊ 1⤋