Factor Completely:
12w^4 - 57w^3 -15w^2
Thanks so much!
2007-05-07 19:35:18 · 7 answers · asked by ♥Cheerleader♥ 1 in Science & Mathematics ➔ Mathematics
3w^2(w-5)(4w+1)
2007-05-07 19:38:54 · answer #1 · answered by Anonymous · 0⤊ 1⤋
= w^2(12w^2 - 57w - 15)
= 3w^2(4w^2 - 19w - 5)
= 3w^2(4w + 1)(w - 5)
2007-05-08 02:40:48 · answer #2 · answered by Lobster 4 · 0⤊ 0⤋
3w^2 (4w^2-19w-5)
wow. i think u hav lots of account,
questions like that
2007-05-08 08:46:09 · answer #3 · answered by enhein 1 · 0⤊ 0⤋
3w^2(4w^2-19w-5)=
3w^2(4w+1)(w-5)
w1=0, w2=-1/4, w3=5
that's all
2007-05-08 02:42:29 · answer #4 · answered by someguy 2 · 0⤊ 0⤋
= 3w² (4w² - 19w - 5)
= 3w².(4w + 1).(w - 5)
2007-05-08 03:41:56 · answer #5 · answered by Como 7 · 0⤊ 0⤋
Hmm...it's fairly simple to figure out. I've got it all written down on this paper.
No, I won't share it with you! That's cheating!!!
2007-05-08 02:39:41 · answer #6 · answered by John Jacob Jingleheimer Schmidt 1 · 1⤊ 1⤋
take w^2 common and then just use the quadratic equation......
all of your previous question are about math questions.....have u ever done any assignment by yourself....
2007-05-08 02:42:20 · answer #7 · answered by hasham1983 3 · 0⤊ 1⤋