(5c - 8)^2 - (2c - 5)^2
thanks for the help
2007-09-09 01:06:01 · 8 answers · asked by K Rose 3 in Science & Mathematics ➔ Mathematics
Let x = 5c - 8 and y = 2c - 5
x ² - y ² = (x - y) (x + y)
= ( 5c - 8 - 2c + 5 ) ( 5c - 8 + 2c - 5 )
= ( 3c - 3 )( 7c - 13 )
= 3 ( c - 1 ) ( 7c - 13 )
2007-09-12 22:02:48 · answer #1 · answered by Como 7 · 1⤊ 0⤋
difference of squares
a^2 - b^2 = (a - b)(a + b)
[5c - 8 - (2c - 5)][5c - 8 + 2c - 5]
(5c - 8 - 2c + 5)(7c - 13)
(3c -3)(7c - 13)
2007-09-09 08:15:22 · answer #2 · answered by Anonymous · 1⤊ 0⤋
DOPS (Difference of 2 perfect squares)
and its factorising, not simplyfying
=[5c-8+(2c-5)][5c-8-(2c-5)]
=(5c-8+2c-5)(5c-8-2c+5)
=(7c-13)(3c-3)
2007-09-09 08:25:32 · answer #3 · answered by Benan L 1 · 1⤊ 0⤋
(5c - 8)^2 - (2c - 5)^2
=(5c-8+2c-5)(5c-8-2c+5) >>> by a^2 - b^2 = (a+b)(a-b)
=(7c-13)(3c-3)
2007-09-09 08:17:11 · answer #4 · answered by Bored 3 · 1⤊ 0⤋
=(5c - 8)^2 - (2c - 5)^2
= (25c^2 - 40c - 40c + 64) - (4c^2 - 10c - 10c + 25)
= (25c^2 - 80c + 64) - (4c^2 - 20c + 25)
= 25c^2 - 80c + 64 - 4c^2 + 20c - 25
= 21c^2 - 60c + 39
= 3(7c^2 - 20c + 13)
= 3(c - 1)(7c - 13)
2007-09-13 07:39:06 · answer #5 · answered by Jun Agruda 7 · 3⤊ 0⤋
(5c-8)*(5c-8)-(2c-5)*(2c-5) [multiply all in brackets to square terms]
25c^2-40c-40c-64-4c^2-10c-10c-25 [add like terms together]
21c^2-100c-89 [simplified]
2007-09-09 08:15:09 · answer #6 · answered by Goethe's Ghostwriter 7 · 0⤊ 2⤋
(5c - 8)^2 - (2c - 5)^2
= 25c^2 - 80c + 64 - ( 4c^2 - 20c + 25)
= 25c^2 - 80c + 64 - 4c^2 + 20c - 25
=21c^2- 60c - 39
formula
(a -b)^2 = a^2 - 2ab + b^2
2007-09-09 08:12:54 · answer #7 · answered by CHENG 2 · 1⤊ 2⤋
25c^2 -80c +64 - 4c^2 +20c - 25
21c^2 -60c +39
3(7c^2 - 20c +13)
3(c-1)(7c-13)
2007-09-09 08:18:52 · answer #8 · answered by Anonymous · 1⤊ 1⤋