2007-09-09 11:55:22 · 8 answers · asked by ARSADIA W 1 in Science & Mathematics ➔ Mathematics
Assuming by "solve" you mean give the values of x for which the formula equals zero, any time one of the terms is zero, the whole thing is zero, so x= 0, x = -4 or x =4 are all solutions.
2007-09-09 11:59:31 · answer #1 · answered by spongeworthy_us 6 · 0⤊ 0⤋
the easiest and fastest way to solve this problem is when remember the formula of a^2 - b^2 = (a+b)(a-b)
so you got x^2-16 right
then time x^2-16 with 3x then you got
3x^3-48x
2007-09-09 19:04:10 · answer #2 · answered by cybersoro 1 · 0⤊ 0⤋
it's not an equation (big giveaway: there's no equals sign), so there's nothing to "solve"
perhaps you want to expand. just multiply the factors.
but note that the last two factors are special and when multiplied will give the difference of two squares:
x^2-16
now multiply this by the first factor (distribute!)
3x*(x^2-16)
which is:
3x^3 - 48x
2007-09-09 18:59:46 · answer #3 · answered by Anonymous · 1⤊ 0⤋
x = 0, -4, 4
2007-09-09 19:02:33 · answer #4 · answered by krysteven 4 · 0⤊ 0⤋
Since it is not an equation, there is no solution
Combine terms.
(3x)(x+4)(x-4)
(3x)(x² - 16)
3x³ - 48x
.
2007-09-09 19:00:03 · answer #5 · answered by Robert L 7 · 2⤊ 0⤋
next step would be
(3x)(x^2-4x+4x-16)
(3x)(x^2-16)
3x^3-48x
2007-09-09 19:06:04 · answer #6 · answered by JC 2 · 0⤊ 0⤋
You don't solve that because it is not an equation. Can you put the complete equation in a comment, please?
2007-09-09 18:59:29 · answer #7 · answered by Tom K 6 · 0⤊ 0⤋
5x
i think
2007-09-09 18:58:37 · answer #8 · answered by kellyluvsyou 1 · 0⤊ 0⤋