No.
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3x2 + 4x factors to x(3x + 4)
(3x4 + 2x3 - x2 - x - 6)/(x2 + 1)= 3x2 + 2x - 4 + (-3x - 2)/(x2 + 1)= 3x2 + 2x - 4 - (3x + 2)/(x2 + 1)where the quotient is 3x2 + 2x - 4 and the remainder is -(3x + 2).
(3x + 4)(2x - 1) = 0 3x = -4 x = -4/3 2x = 1 x = 1/2 x = 1/2, -4/3
3x4,1x12 and 6x2
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)