3x2 + 2x - 8 = 3x2 + 6x - 4x - 8 = 3x(x+2) - 4(x+2) = (3x-4)*(x+2)
-(3x - 4)(x - 2)
It is: (-3x-4)(x+2) when factored
While it is possible to factor 3x2 from both of these and get 3x2(4 - 1), it's a lot easier to subtract 3x2 from 12x2 and get 9x2
4x + 2y = -8 → 2y = -4x - 8 → y = -2x - 4
3x2 + 2x - 8 = 3x2 + 6x - 4x - 8 = 3x(x + 2) - 4(x + 2) = (x + 2)(3x - 4).
if you rearrange it, it becomes 4y + 8 = 3x2 -2x +1 4y = 3x2 -2x -7 y= (3x2 -2x -7)/4 which is a parabola
3x2 + 2x - 8 = 3x2 + 6x - 4x - 8 = 3x(x+2) - 4(x+2) = (3x-4)*(x+2)
3x2-2x-8 = (3x+4)(x-2) when factored
3x2 + 2x = 16 ∴ 3x2 + 2x - 16 = 0 ∴ 3x2 - 6x + 8x - 16 = 0 ∴ 3x(x - 2) + 8(x - 2) = 0 ∴ (3x + 8)(x - 2) = 0 ∴ x ∈ {-8/3, 2}
3x2 + 2x - 8 = 0 is a quadratic equation.
if: f(x) = x3 - 4xe-2x Then: f'(x) = 3x2 - [ 4e-2x + 2(4x / -2x) ] = 3x2 - 4e-2x + 4
(3x+4)(x-2) 3x2 -6x+4x-8 3x2-2x-8
-2
3x2-2x-21 = (3x+7)(x-3)
6x-6x=0 so all you have left is -16 and there is nothing to factor. I am guessing you meant 6x2 -6x-16 which is 2(3x2 -2x-8) The factor pairs for 8 are 1,8 and 2,4. So let's try 1, 8. We know one of them has to be negative. We see it does not work and when we try 2 and 4 they do not work also. So we cannot factor (3x2 -2x-8) over the integers and we call it prime. The final answer is 2(3x2 -2x-8)
3x2 + 2x - 21 = (3x - 7) (x + 3)