+386
(2x^2 + 3x - 2 ) / ( 6x+12)
can this be simplified ?
I thought of:
(x^2 + 3/2x - 2) / (x+6)
Should I leave it in the original formulation?
Thanks!
Simplify the following:
(2 x^2+3 x-2)/(6 x+12)
Factor 6 out of 6 x+12:
(2 x^2+3 x-2)/6 (x+2)
Factor the quadratic 2 x^2+3 x-2. The coefficient of x^2 is 2 and the constant term is -2. The product of 2 and -2 is -4. The factors of -4 which sum to 3 are -1 and 4. So 2 x^2+3 x-2 = 2 x^2+4 x-x-2 = 2 (2 x-1)+x (2 x-1):
2 (2 x-1)+x (2 x-1)/(6 (x+2))
Factor 2 x-1 from 2 (2 x-1)+x (2 x-1):
(2 x-1) (x+2)/(6 (x+2))
((2 x-1) (x+2))/(6 (x+2)) = (x+2)/(x+2)×(2 x-1)/6 = (2 x-1)/6:
Answer: |(2x - 1)/6
Simplify the following:
(2 x^2+3 x-2)/(6 x+12)
Factor 6 out of 6 x+12:
(2 x^2+3 x-2)/6 (x+2)
Factor the quadratic 2 x^2+3 x-2. The coefficient of x^2 is 2 and the constant term is -2. The product of 2 and -2 is -4. The factors of -4 which sum to 3 are -1 and 4. So 2 x^2+3 x-2 = 2 x^2+4 x-x-2 = 2 (2 x-1)+x (2 x-1):
2 (2 x-1)+x (2 x-1)/(6 (x+2))
Factor 2 x-1 from 2 (2 x-1)+x (2 x-1):
(2 x-1) (x+2)/(6 (x+2))
((2 x-1) (x+2))/(6 (x+2)) = (x+2)/(x+2)×(2 x-1)/6 = (2 x-1)/6:
Answer: |(2x - 1)/6
+129840 (2x^2 + 3x - 2 ) / ( 6x+12)
[(2x - 1) (x +2)] / [ 6 (x + 2) ] the (x + 2) factors "cancel"
(2x - 1) / 6
+386 Thanks a lot guys.
I'm trying to work this out with gauss and it doesn't add up.
2x^2 + 3x - 2 's rationnal roots can be - or + 1, - or + 2.
I try replacing x with all four options but nothing =0. Is this normal?
+386 Hehe well by hand x= -2 gives 0 .
Seems the calculator fails with the parenthesis or something. My bad...
