What is the factored form of 2(x - 2)^2 - (x - 2) - 1 ?
2006-07-18 08:11:29 · 5 answers · asked by zoso_arivolk 1 in Science & Mathematics ➔ Mathematics
2(x - 2)^2 - (x - 2) - 1
think of this as
2x^2 - x - 1
this factors to
(2x + 1)(x - 1)
now replace x with (x - 2) and you get
(2(x - 2) + 1)((x - 2) - 1)
(2x - 4 + 1)(x - 2 - 1)
(2x - 3)(x - 3)
ANS : (2x - 3)(x - 3)
2006-07-18 13:57:48 · answer #1 · answered by Sherman81 6 · 2⤊ 1⤋
I will get you started. Let w = x-2, so you have 2w^2-w-1, you should be able to factor this. Once you have factored the w expression substitute w = x-2 and then simplify.
2006-07-18 15:21:02 · answer #2 · answered by raz 5 · 0⤊ 0⤋
I'm going to say ...
[2(x-2)+1][(x-2)-1]
or
[2x-3][x-3]
2006-07-18 15:24:40 · answer #3 · answered by Daniel R 2 · 0⤊ 0⤋
2(x-2)^2 - (x-2) - 1
=2(x-2)^2 - 2(x-2) + (x-2) -1
= 2(x-2){ (x-2)-1} +1{(x-2)-1}
= {(x-2)-1} {2(x-2)+1}
= (x-2-1)(2x-4+1)
= (x-3) (2x-3)
2006-07-18 15:22:42 · answer #4 · answered by flori 4 · 0⤊ 0⤋
2(x - 2)^2 - (x - 2) - 1
Let z = x - 2
2z^2 -z - 1
(2z + 1)(z - 1)
Replace z
[2(x - 2) + 1][(x - 2) - 1)]
[2x - 4 + 1][x - 2 - 1]
[2x - 3][x - 3]
2006-07-18 16:34:01 · answer #5 · answered by kindricko 7 · 0⤊ 0⤋