2007-05-06 16:42:41 · 8 answers · asked by senora_tt 1 in Science & Mathematics ➔ Mathematics
2x^2-7x-4 = 2x^-8x+x-4 =2x(x-4)+1(x-4) = (2x+1)(x-4)
2007-05-06 16:55:30 · answer #1 · answered by rkbaqaya 5 · 0⤊ 0⤋
You'll factor it into two sets of parenthesis. You have 2x^2 to begin with, so it will probably be 2x in one set at the front and x in the front of the other.
(x plus or minus ?)(2x plus or minus ?)
We'll figure out what the two question marks are and what the two signs should be (plus or minus)
You know your last term needs to be a 4, so you will either have 2 in each or 4 in one, 1 in the other.
Let's try 1 and 4.
(2x plus or minus 1)(x plus or minus 4)
Since the 4 needs to be negative, the only way you can get a negative number when multipying is to multiply a negative and a positive, so one sign will be positive and one will be negative.
(2x - 1)(x - 4) When you FOIL (First, Outside, Inside, Last), and combine like terms, you'll get the answer you started with.
2007-05-06 23:53:00 · answer #2 · answered by jkicker 3 · 1⤊ 0⤋
(1,-4) and (2,1)
from x^2-y^2 = (x+y)(x-y)
x - values 1 x 2 = 2 ; 2x^2
y-vaules -4 x 1 = -4
check if factor is equal to -7x by multiplying -4 and 2 plus 1*1
=-7
therefore factors are
(x-4) (2x+1)
2007-05-06 23:47:30 · answer #3 · answered by Maan_palaboy 2 · 0⤊ 0⤋
2x^2 -7x - 4 = (2*x + 1) * (x - 4)
2007-05-06 23:49:50 · answer #4 · answered by Baselek 2 · 0⤊ 0⤋
(2x + 1).(x - 4)
Check
2x.(x - 4) + 1.(x - 4)
= 2x² - 8x + x - 4
= 2x² - 7x - 4 as required.
2007-05-07 06:14:35 · answer #5 · answered by Como 7 · 0⤊ 0⤋
(x+(1/2))(x-4)
2007-05-06 23:51:52 · answer #6 · answered by hockeyjockey0987 2 · 0⤊ 0⤋
(2x + 1)(x - 4)
2007-05-06 23:47:30 · answer #7 · answered by Jason P 4 · 0⤊ 0⤋
(2x+1)(x-4)
2007-05-06 23:48:03 · answer #8 · answered by Anonymous · 0⤊ 0⤋