# How do you use the rational roots theorem to find all possible zeros of f(x) = x^4 + 3x^3 - 4x^2 + 5x -12?

Jun 13, 2016

$f \left(x\right)$ has no rational zeros.

#### Explanation:

Given:

$f \left(x\right) = {x}^{4} + 3 {x}^{3} - 4 {x}^{2} + 5 x - 12$

By the rational root theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 12$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 6$, $\pm 12$

None of these work, so $f \left(x\right)$ has no rational zeros, but in the process of checking, we find:

$f \left(- 6\right) = 462 > 0$

$f \left(- 4\right) = - 32 < 0$

$f \left(1\right) = - 7 < 0$

$f \left(2\right) = 22 > 0$

So $f \left(x\right)$ has a couple of Real irrational zeros in $\left(- 6 , - 4\right)$ and $\left(1 , 2\right)$

Here's a graph of $f \frac{x}{10}$ ...

graph{(x^4+3x^3-4x^2+5x-12)/10 [-10.705, 9.295, -6.56, 3.44]}

This particular quartic is messy to solve algebraically.

You can use a numerical method such as Durand Kerner, to find approximations to the zeros of our quartic:

$- 4.3358$

$1.4332$

$- 0.04871 \pm 1.38880 i$

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Footnote

Interestingly, the similar looking quartic ${x}^{4} + 3 {x}^{3} - 4 {x}^{2} + 6 x - 12$ is much easier to solve, since it factors as:

${x}^{4} + 3 {x}^{3} - 4 {x}^{2} + 6 x - 12 = \left({x}^{2} + 2\right) \left({x}^{2} + 3 x - 6\right)$

Still no rational factors, but much, much easier.