Factor completely
-3t^3 + r^2-6r
2006-10-12 19:30:54 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
if all r's then:
-3r^3+r^2-6r
= -r(3r^2-r+6)
2006-10-13 03:35:11 · answer #1 · answered by SP 1 · 0⤊ 0⤋
Well my friend, there seems to be an error in the problem
Either there are all 't' or there are all 'r'
However, with what you've provided, let me solve d same for you
-3t^3 + r^2 - 6r
= -3t^3 + r(r-6)
2006-10-13 02:36:25 · answer #2 · answered by aazib_1 3 · 0⤊ 0⤋
r^2 - 6r - 3t^3: using quadratic formula,
r = (6 +/- sqrt(36+12t^3))/2 = 3 +/- sqrt(9+3t^3)
It's not a nice factorization, but it's
(r - 3 - sqrt(9+3^3))(r - 3 + sqrt(9+3t^3))
2006-10-13 02:35:10 · answer #3 · answered by James L 5 · 0⤊ 0⤋
if it is in t and r this cannot be done
if it is -3t^3+t^2-6t
then take -t as common we get -t(3t^2-t + 6)
2006-10-13 02:35:38 · answer #4 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
I'm quite sure there is an error in the problem.
maybe the "t" should be "r"
then you can take "r" out common and it will become a quadratic equation
r(-3r^2+r-6)
2006-10-13 03:35:54 · answer #5 · answered by confused 2 · 0⤊ 0⤋
Assuming that the "t" is supposed to be an "r", simply factor out one of the "r's" and you're left with a quadratic, which I hope you can solve on your own (hint: complete the square or use the AC method).
2006-10-13 02:35:23 · answer #6 · answered by ohmneo 3 · 0⤊ 0⤋