I'm lost!
I need help simplifying
3(x^2)^3 (-2x^4)^2
Thanks!!!!!!
2006-12-14 19:02:57 · 7 answers · asked by Thomas and M 1 in Science & Mathematics ➔ Mathematics
3(x^2)^3 (-2x^4)^2
3(x^6) (-4x^8)
3x^6-4x^8
2006-12-14 19:07:49 · answer #1 · answered by jamaica 5 · 0⤊ 0⤋
12 (x^14)
-2 is constant so you can rewrite (-2x^4)^2 as (-2)^2 * (x^4)^2
so now the problem looks like:
3 * (x^2)^3 * (-2)^2 * (x^4)^2
pull out the constants so you have 3*(-2)^2 = 12
then evaluate the exponents (x^2)^3 = x^6 and (x^4)^2 = x^8
x^6 * x^8 = x^14
so the simplified answer is 12 (x^14)
2006-12-14 19:18:25 · answer #2 · answered by kky1313131313 4 · 0⤊ 0⤋
3(x^2)^3(-2x^4)^2
= 12 (x^6)(x^8)
= 12 x^14
2006-12-14 20:19:02 · answer #3 · answered by Srinivas c 2 · 0⤊ 0⤋
3(x^2)^3 (-2x^4)^2
Do all exponents
3(x^6)+4x^8
Then multiply
3x^6+4x^8
2006-12-14 19:06:44 · answer #4 · answered by endless_h8 2 · 0⤊ 0⤋
12x^14
2006-12-14 19:07:30 · answer #5 · answered by y-x 1 · 0⤊ 0⤋
-12x^14
2006-12-14 19:05:45 · answer #6 · answered by Anonymous · 0⤊ 0⤋
3.x^6.4x^8
= 12x^14
remember two rules:
(x^n)^m = x^(mn)
x^m.x^n = x^(m+n)
2006-12-14 19:05:27 · answer #7 · answered by alia_vahed 3 · 0⤊ 0⤋