2007-10-05 02:39:14 · 4 answers · asked by . 1 in Science & Mathematics ➔ Mathematics
12y^2-16y-9y+12
=(4y-3)(3y-4)
2007-10-05 02:43:57 · answer #1 · answered by cooldude_raj07 2 · 0⤊ 0⤋
12y² - 25y + 12 = 0
The middle term is - 25y
Find the sum of the middle term
Multiply the first term 12 times the last term 12 equals 144 and factor
Factors of 144
1 x 144
2 x 72
3 x 48
4 x 36
6 x 24
8 x 18
9 x 16. . .â. .use these factors
12 x 12
- 16 and - 9 satisfy the sum of the middle term
insert - 16y and - 9y into the equation
12y² - 25y + 12 = 0
12y² - 16y - 9y + 12 = 0
Group factor
(12y² - 16y) - ( 9y + 12) = 0
4y(3y - 4) - 3(3y - 4) = 0
(4y - 3)(3y - 4) = 0
- - - - - - - - -s-
2007-10-05 10:13:57 · answer #2 · answered by SAMUEL D 7 · 0⤊ 0⤋
12y^2 - 25y + 12 can be rewritten as
12y^2 - 16 y -9y +12
4y(3y -4) - 3(3y-4) = (4y-3) (3y-4).. answer
2007-10-05 10:11:51 · answer #3 · answered by Metallurgist 2 · 0⤊ 0⤋
12(x-.75)(x-1.33333)
2007-10-05 09:52:51 · answer #4 · answered by Anonymous · 0⤊ 1⤋