I need help factoring this polynomial 16 a ^4 - 81 b ^4 can anyone help? Thanks
2007-06-15 20:17:38 · 10 answers · asked by ambi565 2 in Science & Mathematics ➔ Mathematics
16 a ^4 - 81 b ^4
( 4a^2 + 9b^2 )*(4a^2 - 9b^2)
(2a + 3b)*(2a - 3b)*(2a + 3b)*(2a - 3b)
I haven't done this kind of problem in a very long time! See "gp4rts" for why.
2007-06-15 20:30:40 · answer #1 · answered by sheik_sebir 4 · 1⤊ 0⤋
16*a^4 and 81*b^4 are perfect squares. The difference of perfect squares is solved by the relation
x^2 - y^2 = (x + y)*(x - y).
Apply this formula to your problem. Note that when you do this, one of the factors will again be a difference of perfect squares, so you have to do it again.
2007-06-16 03:22:43 · answer #2 · answered by gp4rts 7 · 1⤊ 0⤋
the given polynomial can be written as (4a^2)^2 - (9b^2)^2. Now apply the formula,
(x^2-y^2) = (x-y)(x+y)
here x = 4a^2 and y = 9b^2
Therefore the polynomial can be factored as
(4a^2 - 9b^2)(4a^2+9b^2)
2007-06-16 03:30:43 · answer #3 · answered by csubbu 1 · 0⤊ 0⤋
16 a ^4 - 81 b ^4 =
(4a^2 - 9b^2) (4a^2 + 9b^2) =
(2a - 3b) (2a + 3b) (4a^2 + 9b^2) Final Answer
2007-06-16 03:45:36 · answer #4 · answered by Matt D 6 · 0⤊ 0⤋
16a^4 - 81b^4 =
(4a^2)^2 - (9b^2)^2 = (Difference of square)
(4a^2 + 9b^2) (4a^2 - 9b^2) =
(4a^2 + 9b^2) ( 2a + 3b)(2a - 3b) ==> the answer
2007-06-16 03:30:17 · answer #5 · answered by detektibgapo 5 · 0⤊ 0⤋
never did alebra but if i understand it the answer would be something like 65 c^4....probably wrong but i thought i might give it a go lol
2007-06-16 03:22:50 · answer #6 · answered by hawk_darkknight2000 2 · 0⤊ 0⤋
= (4a² - 9 b²).(4a² + 9b²)
= (2a - 3b).(2a + 3b).(4a² + 9b²)
2007-06-17 02:36:43 · answer #7 · answered by Como 7 · 0⤊ 0⤋
(4a^2-9b^2)^2
2007-06-16 03:24:34 · answer #8 · answered by Anonymous · 0⤊ 0⤋
(2a-3b)^2 (2a+3b)^2
2007-06-16 03:48:56 · answer #9 · answered by konstipashen 5 · 0⤊ 0⤋
i dont have any idea good luck
2007-06-16 03:20:09 · answer #10 · answered by joejoe 1 · 1⤊ 0⤋