factor the poloynomial. this is far as i have gotten:
49x^4-16a^2
(7x^2+4a)(7x^2-4a)
(7x^2+4a)(7x+2a)(7x-2a)...this is where i am stuck
2007-08-27 08:03:14 · 8 answers · asked by arod_69 1 in Science & Mathematics ➔ Mathematics
Actually your last factorization doesn't work because the a isn't squared
The 7x²+4a is not factorable over reals
the 7x²-4a factors as (√7 x-2√a)(√7 x+2√a)
2007-08-27 08:14:12 · answer #1 · answered by chasrmck 6 · 0⤊ 2⤋
You correctly have (7x^2+4a)(7x^2-4a), but to use the "difference of squares" on the last parenthesis, everything in it needs to be "square-rooted," including BOTH the ' 7 ' AND the ' a.'
So you'll get:
(7x^2 + 4a)(sqrt7 x + 2 sqrta)(sqrt7 x - 2 sqrta), that is
(7x^2 + 4a)(â7 x + 2 âa)(â7 x - 2 âa).
Live long and prosper.
2007-08-27 15:14:13 · answer #2 · answered by Dr Spock 6 · 0⤊ 0⤋
49x^4-16a^2
(7x^2+4a)(7x^2-4a) Answer
u need not go beyond this as sq roots will be involved then.
your last line is in-correct
if at all u need to go to that step, it would read as :
(7x^2+4a)(â7 x-2âa)(â7 x+2âa)
2007-08-27 16:12:01 · answer #3 · answered by kapilbansalagra 4 · 0⤊ 1⤋
Your last factor is incorrect because it would sum to -4a^2.
The second step is as far as you can go.
2007-08-27 15:22:23 · answer #4 · answered by thehazmater 2 · 0⤊ 0⤋
(7x^2+4a)(7x^2-4a) = (7x^2-4a)^2
2007-08-27 15:14:11 · answer #5 · answered by cool j 2 · 0⤊ 1⤋
(7x² - 4a)(7x² + 4a) is OK
2007-08-31 07:57:49 · answer #6 · answered by Como 7 · 1⤊ 0⤋
=(7x²-4a)²
2007-08-27 15:14:30 · answer #7 · answered by Anonymous · 0⤊ 1⤋
ur answer is 100% correct
just a mistake in the last line
u have written this is where i am 'stuck'
the sentence should be this is where i solved it
:-):-)
2007-08-27 15:15:48 · answer #8 · answered by ? 2 · 0⤊ 1⤋