1, 8, -5 and 4
One could argue that 4 is not a coefficient, but a term of it's own. On the other hand, if you follow the pattern in the polynomial, you could argue that it's a coefficient of x0.
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x3 + x2 - 6x + 4 = (x - 1)(x2 + 2x - 4)
x3 + 4x2 + 6x + 24 = (x2 + 6)(x + 4)
To factorise x3 + 5x2 - 16x - 80 I note that -80 = 5 x -16 and 5 & -16 are the coefficients of x2 and x. Thus I have: x3 + 5x2 - 16x - 80 = (x + 5)(x2 - 16) and the second term is a difference of 2 squares, meaning I have: x3 + 5x2 - 16x - 80 = (x + 5)(x + 4)(x - 4)
x3 + 12x2 + 48x + 64= (x + 4)(x2 + 8x + 16)= (x + 4)(x + 4)(x + 4)= (x + 4)3
x3 - 3x2 + 4 Since the coefficients of the odd powers of x (=1) is the same as the sum of the even powers (-3+4=1), then x = -1 must be a root. That is to say, (x + 1) is a factor. So you can rewrite the expression as x3 + x2 - 4x2 - 4x + 4x + 4 = x2(x + 1) - 4(x + 1) + 4(x + 1) = (x + 1)*(x2 - 4x + 4) = (x + 1)*(x - 2)2