# What are all the possible rational zeros for f(x)=4x^4-16x^3+12x-30 and how do you find all zeros?

Jan 8, 2017

The "possible" rational zeros are:

$\pm \frac{1}{2} , \pm 1 , \pm \frac{3}{2} , \pm \frac{5}{2} , \pm 3 , \pm 5 , \pm 15$

but this quartic has two irrational and two complex zeros.

#### Explanation:

Note that all of the coefficients of $f \left(x\right)$ are divisible by $2$, so separate that off as a factor before applying the rational roots theorem:

$f \left(x\right) = 4 {x}^{4} - 16 {x}^{3} + 12 x - 30 = 2 \left(2 {x}^{4} - 8 {x}^{3} + 6 x - 15\right)$

By the rational roots theorem, any rational zeros of $2 {x}^{4} - 8 {x}^{3} + 6 x - 15$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 15$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{2} , \pm 1 , \pm \frac{3}{2} , \pm \frac{5}{2} , \pm 3 , \pm 5 , \pm 15$

Note that:

$2 \left({\textcolor{b l u e}{5}}^{4}\right) = 1250 > 1045 = 8 \left({\textcolor{b l u e}{5}}^{3}\right) + 6 \left(\textcolor{b l u e}{5}\right) + 15$

Hence any zeros lie strictly inside the circle $\left\mid x \right\mid = 5$ in the complex plane.

So we can discard $\pm 5$ and $\pm 15$ as possibilities.

To cut a long story short, none of the "possible" rational zeros are zeros of this quartic.

It is a typical "nasty" quartic with complicated irrational zeros.

It is possible to solve algebraically, but much simpler to find numerical approximations to the zeros using a method such as Durand-Kerner.

${x}_{1} \approx - 1.2905$

${x}_{2} \approx 3.9293$

${x}_{3 , 4} \approx 0.68061 \pm 1.00785 i$

See https://socratic.org/s/aB9Ee9wQ for another example.