how do you factor a problem that is like this:
81p^4 - 18p^2 + 1 = 0
please explain
thank you
2007-08-30 14:10:20 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
thanks everyone!
2007-08-30 14:23:49 · update #1
find common denominator (9p^2) and take it out
=(9p^2)(9p^2) - 2 +1=0
then add= (9p^2)(9p^2)-1=0 then add 1 to both sides
(9p^2)(9p^2)=1 then find what p is if needed to go further
p=0,1/3,-1/3
2007-08-30 14:21:02 · answer #1 · answered by musique 2 · 0⤊ 0⤋
81p^4 -- 18p^2 + 1 = 0
=> (9p^2 -- 1)^2 = 0
=> 9p^2 -- 1 = 0
=> (3p + 1)(3p -- 1) = 0
=> p = --1/3, 1/3
2007-08-30 14:23:27 · answer #2 · answered by sv 7 · 0⤊ 0⤋
(9p^2 - 1)(9p^2 - 1) or (9p^2 -1)^2 Can't explain. It takes a lot of practice doing homework problems to see how to do these.
2007-08-30 14:26:37 · answer #3 · answered by Renaissance Man 5 · 0⤊ 0⤋
Anything times 0 equals 0
2007-08-30 14:16:46 · answer #4 · answered by stunning_lie67 1 · 0⤊ 0⤋
(9p^2 - 1) ^2 = 0
2007-08-30 14:16:54 · answer #5 · answered by Anonymous · 0⤊ 0⤋
(9p² - 1) (9p² - 1) = 0
p² = 1 / 9
p = ± (1 / 3)
2007-08-30 21:49:54 · answer #6 · answered by Como 7 · 1⤊ 0⤋