1) 4y^4-25x^2
2) y^3-8
thanks
2006-10-29 12:06:38 · 8 answers · asked by Anonomous101 1 in Science & Mathematics ➔ Mathematics
1) 4y^4-25x^2 = (2y^2 - 5x) ( 2y^2+5x)
2) y^3-8 =(y-2)(y^2+2y+4)
2006-10-29 12:17:55 · answer #1 · answered by Anonymous · 0⤊ 2⤋
1.)
4y^4 - 25x^2 = (2y^2 - 5x)(2y^2 + 5x)
2.)
y^3 - 8 = (y - 2)(y^2 + 2y + 4)
2006-10-29 15:12:28 · answer #2 · answered by Sherman81 6 · 1⤊ 0⤋
1) 4y^2-25x^2
= (2y^2 + 5x)(2y^2 - 5x)
2) y^3 - 8
= (y-2)(y^2 + 2y + 4)
2006-10-29 13:30:20 · answer #3 · answered by Danny_V 1 · 1⤊ 0⤋
4y^4 - 25x^2 =
(2y^2 - 5x) (2y^2+5x)_
2006-10-29 12:17:34 · answer #4 · answered by Steve A 7 · 0⤊ 1⤋
1 = (2y^2 - 5x)(2y^2 + 5x)
The difference of 2 squares (A^2-B^2) can be factored into (A-B)(A+B)
I don't recall the rules for difference of cubes at the moment
~ ⥠~
2006-10-29 12:16:14 · answer #5 · answered by I ♥ AUG 6 · 2⤊ 1⤋
4y⁴ - 25x² =
(2x² + 5x)(2x² - 5x)
- - - - - - - - - - - - - -
FOIL Method
(2y² + 5x)(2y² - 5x) = 4y⁴ + 10xy² - 10xy ² - 25x² =
4y⁴- 25y²
- - - - - - -s-
2006-10-29 13:13:13 · answer #6 · answered by SAMUEL D 7 · 0⤊ 0⤋
1) =(2y^2-5x)(2y^2+5x)
2)=(y-2)(y^2+2y+4)
2006-10-29 12:18:54 · answer #7 · answered by peterwan1982 2 · 0⤊ 0⤋
1) is a difference of squares.
2) is a difference of cubes.
Good luck!
2006-10-29 12:10:28 · answer #8 · answered by Anonymous · 1⤊ 1⤋