3x2 + 2x - 21 = (3x - 7) (x + 3)
3x3 - 3x2 + 2x - 2 = 3x2(x - 1) + 2(x - 1) = (3x2 + 2)(x - 1)
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if you rearrange it, it becomes 4y + 8 = 3x2 -2x +1 4y = 3x2 -2x -7 y= (3x2 -2x -7)/4 which is a parabola
(3x - 1)(x + 1) or (3x + 1)(x - 1) assuming the 1 is negative.
3x2-2x-21 = (3x+7)(x-3)
3x2 + 2x - 21 = (3x - 7) (x + 3)
x(3x-2)
3(x2 - 2x + 3)
It is written in descending order.
3X2 + 6X + 9 3(X2 + 2 + 3) -----------------------3 is a common factor of all terms
3x2 + 2x - 8 = 3x2 + 6x - 4x - 8 = 3x(x+2) - 4(x+2) = (3x-4)*(x+2)
3x3 - 3x2 + 2x - 2 = 3x2(x - 1) + 2(x - 1) = (3x2 + 2)(x - 1)
(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)
if: f(x) = x3 - 4xe-2x Then: f'(x) = 3x2 - [ 4e-2x + 2(4x / -2x) ] = 3x2 - 4e-2x + 4
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