12y^3 + 28y^2 + 8y
I get GCF 4 = (4y^3 + 7y^2 + 2y)
( + ) ( + )
2 and 1 2 and 1 (factors)of the first and last
so (2 + 1) ( 2 + 1) Don't equal middle term.
I don't think I am doing this right or maybe this trinomial is prime?
2007-06-11 13:24:27 · 6 answers · asked by lexisayn 1 in Education & Reference ➔ Homework Help
12y^3 + 28y^2 + 8y
4y(3y² + 7y + 2)
4y(3y + 1)(x + 2)
.
2007-06-11 13:32:19 · answer #1 · answered by Robert L 7 · 1⤊ 0⤋
I came up with this:
4y(3y^2 + 7y + 2)
I know you can do factoring beyond that but I'm stumped. Been a little too long since I've taken Algebra.
EDIT:
Okay, to the bozo who gave myself and another a thumbs down, good grief. At least we TRIED to help the girl. Damn. Excuse me for not being a math genius.
2007-06-11 13:30:05 · answer #2 · answered by Anonymous · 0⤊ 1⤋
Try this...
4y (3y^2 + 7y + 2) = 4y (3y + 1)(y + 2)
2007-06-11 13:29:15 · answer #3 · answered by Cindy 3 · 0⤊ 1⤋
GCF is 4y
4y(3y(squared) + 7y +2)
so...i can't figure the rest.
2007-06-11 13:33:28 · answer #4 · answered by Anonymous · 0⤊ 0⤋
factorising n^2 - 16n + sixty 4, that's written as (n)^2 - 2 * 8 n + (8)^2 using the identification a^2 - 2*a*b + b^2 = (a-b)^2 =>(n-8)^2 = (n-8)(n-8) you additionally can call n^2 - 16n + sixty 4, and a million to be its element. wish it facilitates
2016-12-12 18:29:06 · answer #5 · answered by ? 4 · 0⤊ 0⤋
first simplify then do whatever you did and you will get your anwser
2007-06-11 13:31:42 · answer #6 · answered by syed m 2 · 0⤊ 1⤋