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

1 Answer
Feb 20, 2018

Answer:

#-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)#