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

Feb 20, 2018

$- \frac{1}{6 {x}^{3} + 3 {x}^{2} + 5 x - 4} = - \frac{1}{2 x - 1} \cdot \frac{1}{3 {x}^{2} + 3 x + 4}$

#### Explanation:

I wonder whether the question meant to factor:

$- \frac{1}{6 {x}^{3} + 3 {x}^{2} + 5 x - 4}$

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 $6 {x}^{3} + 3 {x}^{2} + 5 x - 4$

By the rational roots theorem, the only possible rational zeros of $6 {x}^{3} + 3 {x}^{2} + 5 x - 4$ are expressible in the form $\frac{p}{q}$ 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:

$\pm \frac{1}{6} , \pm \frac{1}{3} , \pm \frac{1}{2} , \pm \frac{2}{3} , \pm 1 , \pm \frac{4}{3} , \pm 2 , \pm 4$

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:

$6 {\left(\textcolor{b l u e}{\frac{1}{2}}\right)}^{3} + 3 {\left(\textcolor{b l u e}{\frac{1}{2}}\right)}^{2} + 5 \left(\textcolor{b l u e}{\frac{1}{2}}\right) - 4 = \frac{3}{4} + \frac{3}{4} + \frac{5}{2} - 4 = 0$

So $x = \frac{1}{2}$ is a zero and $\left(2 x - 1\right)$ a factor:

$6 {x}^{3} + 3 {x}^{2} + 5 x - 4 = \left(2 x - 1\right) \left(3 {x}^{2} + 3 x + 4\right)$

Note that the discriminant of the remaining quadratic factor is negative:

${\left(\textcolor{b l u e}{3}\right)}^{2} - 4 \left(\textcolor{b l u e}{3}\right) \left(\textcolor{b l u e}{4}\right) = 9 - 48 = - 39$

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:

$- \frac{1}{6 {x}^{3} + 3 {x}^{2} + 5 x - 4} = - \frac{1}{2 x - 1} \cdot \frac{1}{3 {x}^{2} + 3 x + 4}$