Question

How do you factor `-1/6x^3+3x^2+5x-4`?

Answer 1

`-1/(6x^3+3x^2+5x-4) = -1/(2x-1) * 1/(3x^2+3x+4)`

Explanation:

I wonder whether the question meant to factor:

`-1/(6x^3+3x^2+5x-4)`

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 `6x^3+3x^2+5x-4`

By the rational roots theorem, the only possible rational zeros of `6x^3+3x^2+5x-4` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-4` and `q` a divisor of the coefficient `6` of the leading term.

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 `+ + + -`. With one change of signs, Descartes' Rule of Signs tells us that this cubic has exactly one positive real zero.

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`

So `x=1/2` is a zero and `(2x-1)` a factor:

`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)`