# How do you find all rational roots for 3x^4 - 2x^3 + 8x^2 - 6x - 3 = 0?

May 30, 2016

The rational root theorem can point out the possibilities, but in this particular case there's a shortcut to the rational zeros:

$x = 1$ and $x = - \frac{1}{3}$

#### Explanation:

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

By the rational root theorem, any rational zero of $f \left(x\right)$ is expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divior of the constant term $- 3$ and $q$ a divisor of the coefficient $3$ of the leading term.

So the only possible rational zeros are:

$\pm \frac{1}{3}$, $\pm 1$, $\pm 3$

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Actually we can proceed more directly to find all the zeros as follows:

Note that the sum of the coefficients of $f \left(x\right)$ is zero. That is:

$3 - 2 + 8 - 6 - 3 = 0$

Hence $f \left(1\right) = 0$ and $\left(x - 1\right)$ is a factor:

$3 {x}^{4} - 2 {x}^{3} + 8 {x}^{2} - 6 x - 3 = \left(x - 1\right) \left(3 {x}^{3} + {x}^{2} + 9 x + 3\right)$

The remaining cubic can be factored by grouping as follows:

$3 {x}^{3} + {x}^{2} + 9 x + 3$

$= \left(3 {x}^{3} + {x}^{2}\right) + \left(9 x + 3\right)$

$= {x}^{2} \left(3 x + 1\right) + 3 \left(3 x + 1\right)$

$= \left({x}^{2} + 3\right) \left(3 x + 1\right)$

So another rational zero is $x = - \frac{1}{3}$

The remaining quadratic has no Real zeros, rational or irrational since ${x}^{2} + 3 \ge 3 > 0$ for any Real value of $x$.

It does have Complex zeros $x = \pm \sqrt{3} i$