How do you factor #-1/6x^3+3x^2+5x-4#?
I wonder whether the question meant to factor:
since the question as given has three irrational real zeros only expressible in terms of trigonometric expressions or cube roots of complex numbers.
So let's factor
By the rational roots theorem, the only possible rational zeros of
That means that the only possible rational zeros are:
#+-1/6, +-1/3, +-1/2, +-2/3, +-1, +-4/3, +-2, +-4#
In addition note that the pattern of the signs of the coefficients is
So, trying the positive possibilities first we find:
#6(color(blue)(1/2))^3+3(color(blue)(1/2))^2+5(color(blue)(1/2))-4 = 3/4+3/4+5/2-4 = 0#
#6x^3+3x^2+5x-4 = (2x-1)(3x^2+3x+4)#
Note that the discriminant of the remaining quadratic factor is negative:
#(color(blue)(3))^2 - 4(color(blue)(3))(color(blue)(4)) = 9-48 = -39#
So this quadratic only has factors with non-real complex coefficients. Since this question is asked under Algebra, I will stop at the real factorisation to find:
#-1/(6x^3+3x^2+5x-4) = -1/(2x-1) * 1/(3x^2+3x+4)#