^3 is to the third power.
2007-05-27 07:30:39 · 6 answers · asked by Need Help 1 in Science & Mathematics ➔ Mathematics
4(x+6)^3 +17(x-6)^2 -15(x-6)
= 4(x-6)(x+6)(x-6)+17(x-6)^2 -15(x-6)
= (x-6) [ 4(x+6)(x-6)+17(x-6) -15)]
=(x-6) [ (2(x+6)-5) (2(x+6)-3)]
= answer!!
that is the factored answer
hoped it helped.. by the way what grade or course is this for?
2007-05-27 07:52:21 · answer #1 · answered by veena_dracks84 2 · 0⤊ 2⤋
4(x+6)^3 +17(x-6)^2 -15(x-6)
= 4(x-6)(x+6)(x-6)+17(x-6)^2 -15(x-6)
= (x-6) [ 4(x+6)(x-6)+17(x-6) -15)]
=(x-6) [ (2(x+6)-5) (2(x+6)-3)]
= answer!!
that is the factored answer
hoped it helped.. by the way what grade or course is this for?
2007-06-03 03:53:28 · answer #2 · answered by Michael B 2 · 0⤊ 0⤋
uh, so stay in whole numbers for a while im gonna say x = y -3
4(y+3)^3+17(y+3)(y-3)- 15(y-9)
(y+3)[4(y+3)^2+17(y-3)] - 15(y-9)
ok, u go from there
Veena had right idea but picked wrong term
2007-05-27 08:24:23 · answer #3 · answered by Anonymous · 0⤊ 0⤋
4(x+6)^3+17(x-6)^2-15(x-6)
=4(x+6)^3 +(x-6)[(17(x-6)-15)]
= 4(x+6)^3 +(x-6)(17x-123)
Are you sure the 1st term was not 4(x-6)^3?
BTW (x+6)^3 not =(x-6)(x-6)(x+6) as suggested above.
2007-05-27 07:53:20 · answer #4 · answered by ironduke8159 7 · 0⤊ 0⤋
I guess the problem might be
4(x-6)^3-17(x-6)^2+15(x-6)
= (x-6)(x-6-3)[4(x-6)-5)]
= (x-6)(x-9)(4x-29)
2007-05-27 07:55:58 · answer #5 · answered by sahsjing 7 · 0⤊ 0⤋
I tried a few different tricks and got nada. So it looks like it is the brute force method for you ie multiply out the whole thing and collect terms. You will have a trinomial. Use one of the trinomial tricks to get the answer.
2007-05-27 07:42:33 · answer #6 · answered by kellenraid 6 · 0⤊ 2⤋