6y^2 + 17y + 12
2007-07-26 10:41:54 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
6y^2 + 17y + 12 =
(3y + 4)(2y + 3)
2007-07-26 10:47:37 · answer #1 · answered by N E 7 · 1⤊ 0⤋
6y^2 +17y +12
(6y+ )(y + )....?
(2y + )(3y+ )...?
The trick now is to find two numbers that multiply to give 12 and add under the FOIL process to give 17. The FOIL process is a procedure for multiplying factors that contain only two terms, to get a trinomial.
F means multiply First terms
O means multiply Outside terms
I means multiply Inside terms
L means multiply Last terms
(2y + ?)(3y + ?) needs either 12 and 1 as last terms, or 6 and 2, or 4 and 3. I see that 4 and 3 will work IF I write them in as follows:
(2y + 3)(3y + 4). When I FOIL this, I get
F is 6y^2 , O is8y, I is 9y, L is 12. This gives me
6x^2 + 8y +9y +12, or 6x^2 + 17y + 12
Hope this helps
2007-07-26 18:10:10 · answer #2 · answered by Grampedo 7 · 0⤊ 0⤋
6y^2 + 17y + 12 = 0
6y^2 + 9y + 8y + 12 = 0
(6y^2 + 9y) + (8y + 12) = 0
3y (2y + 3) + 4 (2y + 3) = 0
(3y + 4)(2y + 3) = 0
2007-07-26 17:48:44 · answer #3 · answered by dannis8ind 2 · 1⤊ 0⤋
(3y+4)(2y+3)
2007-07-26 17:53:20 · answer #4 · answered by 037 G 6 · 1⤊ 0⤋