2007-01-26 10:22:53 · 6 answers · asked by chris 2 in Science & Mathematics ➔ Mathematics
x^4-81=
(x^2+9)(x^2-9)
(x^2+9)(x+3)(x-3)
2007-01-26 10:30:53 · answer #1 · answered by Bill F 6 · 0⤊ 0⤋
I think you made a mistake in your equation. Is it supposed to be x^4-81=0?
If so
(x-3)(x+3)(x-3)(x+3)=0, I believe. Its been a while and I don't have my Ti-89 to verify
2007-01-26 18:31:06 · answer #2 · answered by K J 1 · 0⤊ 1⤋
a^2-b^2 = (a+b)(a-b)
x^4-81
(x^2)^2-9^2
(x^2+9)(x^2-9)
2007-01-26 18:28:03 · answer #3 · answered by 7 · 0⤊ 1⤋
*You have the differences of squares...
First: write the variables in lowest terms...
(x^2)(x^2) - (9)(9)
Sec: combine one of the 1st set of parenthesis with one from the 2nd set of parenthesis...
(x^2 + 9)(x^2 - 9)
Third: factor the 2nd set of parenthesis > you have the difference of squares....
(x^2 + 9)(x + 3)(x - 3)
2007-01-26 18:40:14 · answer #4 · answered by ♪♥Annie♥♪ 6 · 0⤊ 0⤋
x^4 - 81
= (x^2 - 9) (x^2 +9)
= (x-3)(x+3)(x^2+9)
.... and if you're really nuts ...
= (x-3)(x+3)(x-3i)(x+3i)
2007-01-26 18:29:21 · answer #5 · answered by rpresser 2 · 0⤊ 1⤋
(x^2-9)(x^2+9)
(x-3)(x+3)(x^2-9)
2007-01-26 19:31:54 · answer #6 · answered by Dave aka Spider Monkey 7 · 0⤊ 1⤋