x2/(3x2 - 5x - 2) - [2x/ (3x + 1)][1/(x - 2)]
= x2/(3x2 - 6x + x - 2) - 2x/(3x + 1)(x - 2)
= x2/[3x(x - 2) + (x - 2)] - 2x/(3x + 1)(x - 2)
= x2/(3x + 1)(x - 2) - 2x/(3x + 1)(x - 2)
= (x2 - 2x)/(3x + 1)(x - 2)
= x(x - 2)/(3x + 1)(x - 2)
= x/(3x + 1)
if: f(x) = x3 - 4xe-2x Then: f'(x) = 3x2 - [ 4e-2x + 2(4x / -2x) ] = 3x2 - 4e-2x + 4
3x2+2x-8=0 Delta = 22-4*3*-8= 100 So x = (-2 + sqrt(100))/(2*3) or x = (-2 - sqrt(100))/(2*3) x = 4/3 or x=-2
x=1
3x2 + 2x + 2 = 0 Can not be factored. You can however solve it for x: 3x2 + 2x = -2 x2 + 2x/3 = -2/3 x2 + 2x/3 + 1/9 = -2/3 + 1/9 (x + 1/9)2 = -5/9 x + 1/9 = ±√(-5/9) x = -1/9 ± i√(5/9) x = -1/9 ± i√5 / 3 x = -1/9 ± 3i√5 / 9 x = (-1 ± 3i√5) / 9
4
5x3 - 3x2 + 2x = x*(5x2 - 3x + 2) The quadratic has no real factors.
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 + 3 + x2 + x + 1 = 4x 2+ 3x + 4
If it is 3x2 + 4 = 2x + 4, then you can subtract 4 from both sides, and get 3x2 = 2x, and then x = 2/3
3x2 + 2x - 21 = (3x - 7) (x + 3)
2x+3x2 = x (2+3x)
If you notice, 8x + 4 = 4(2x + 1), but there is an odd coefficient for x2, so there is guaranteed to be some remainder (that is 8x + 4 is not a factor of the polynomial): (24x3 - 5x2 - 48x - 8) ÷ (8x + 4) = 3x2 - 2x - 5 - (x2 - 12)/(8x + 4)
If the original is 2x + 3x2 + 6 + 5x3, you should rearrange the terms in order of exponent:5x3 + 3x2 + 2x + 6
That depends on whether or not 2x is a plus or a minus
6x2-2x+36 = 5x2+10x 6x2-5x2-2x-10x+36 = 0 x2-12x+36 = 0 (x-6)(x-6) = 0 x = 6 or x = 6 It has two equal roots.
3x2 + 2x - 8 = 3x2 + 6x - 4x - 8 = 3x(x + 2) - 4(x + 2) = (x + 2)(3x - 4).
7x2 -4x -5x2 +2x = 2x2 -2x