2007-03-30 08:10:35 · 6 answers · asked by stu_art_27 2 in Science & Mathematics ➔ Mathematics
4a^8 - 28a^4b^5 + 49b^10
Let x = 2a^4 and y = 7b^5
Now you have:
x^2 - 2xy + y^2
(x - y)^2
so, putting the original values back in...
(2a^4 - 7b^5)^2
Just as a note... (2a^4 - 7b^5)^2 yields the same result as (7b^5 - 2a^4)^2, so either are correct.
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Why, thank you, Will M. I'm honored to be called a freak for figuring that problem out. :-)
2007-03-30 08:14:38 · answer #1 · answered by Mathematica 7 · 3⤊ 0⤋
49b^10 + 4a^8 - 28a^4b^5
this should be rewritten as
4a^8 - 28a^4b^5 + 49b^10
(2a^4 - 7b^5)(2a^4 - 7b^5)
or just
(2a^4 - 7b^5)^2
2007-03-30 08:44:26 · answer #2 · answered by Sherman81 6 · 0⤊ 0⤋
=(7b^5-2a^4)^2
2007-03-30 08:31:34 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
(7b^5-2a^4)(7b^5-2a^4) or you could write it like (7b^5-2a^4)^2
2007-03-30 08:21:08 · answer #4 · answered by hrhbg 3 · 1⤊ 0⤋
(7b^5-2a^4)^2
2007-03-30 08:20:51 · answer #5 · answered by LemonButt 3 · 1⤊ 0⤋
the person who answered this before me is a goddamn freak.
2007-03-30 08:18:11 · answer #6 · answered by will m 1 · 1⤊ 2⤋