Factorise this: 25x^2 + 16y^2?
回答 (9)
2018-04-12 7:33 pm
It can't be factorised in the set of real numbers, as the sum of two squares doesn't have any real roots (it is verified only when both x and y are equal to zero).
Not sure if you are meant to solve this with complex numbers.
Not sure if you are meant to solve this with complex numbers.
2018-04-12 7:48 pm
(5x + 4iy)(5x - 4iy) is a possibility.
Or it could just as well be
(4y + 5ix)(4y - 5ix).
Or it could just as well be
(4y + 5ix)(4y - 5ix).
2018-04-12 11:18 pm
25x^2 + 16y^2
= 25x^2 - (-16y^2)
= (5x)^2 - (4iy)^2
= (5x - 4iy)(5x + 4iy)
= 25x^2 - (-16y^2)
= (5x)^2 - (4iy)^2
= (5x - 4iy)(5x + 4iy)
2018-04-12 7:29 pm
Ok
2018-04-12 7:33 pm
(5x+4y)(5x-4y)
2018-04-17 7:54 pm
25x^2 + 16y^2 = (5x)^2 - (4iy)^2 = (5x-4iy)(5x+4iy).
2018-04-13 5:05 pm
25x² + 16y²
= (5x)² + (4y)²
= (5x)² - (-4y)²
= (5x)² - (i 4y)²
= (5x + i4y) (5x - i4y)
= (5x)² + (4y)²
= (5x)² - (-4y)²
= (5x)² - (i 4y)²
= (5x + i4y) (5x - i4y)
2018-04-13 9:45 am
25x^2 + 16y^2 =
(5x + 4iy)(5x - 4iy)
(5x + 4iy)(5x - 4iy)
2018-04-12 8:22 pm
cannot be factored.
Did you mean 25x^2 -16y^2?
Did you mean 25x^2 -16y^2?
收錄日期: 2021-04-24 01:04:57
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20180412112857AAFS1Ho