# How do you find any rational roots for the equation: x^4 + 3x^3 - x^2 - 9x - 6 = 0?

Sep 17, 2016

This quartic has rational roots $1$ and $- 2$

It has irrational zeros $\pm \sqrt{3}$

#### Explanation:

${x}^{4} + 3 {x}^{3} - {x}^{2} - 9 x - 6 = 0$

By the rational roots theorem, any rational roots of this quartic are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 6$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

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

Notice that if we reverse the signs of the coefficients on the terms of odd degree then the sum is $0$. That is:

$1 - 3 - 1 + 9 - 3 = 0$

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

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

Notice that the ratio between the first and second terms of the remaining cubic is the same as that between the third and fourth terms. So this cubic will factor by grouping:

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

$\textcolor{w h i t e}{{x}^{3} + 2 {x}^{2} - 3 x - 6} = {x}^{2} \left(x + 2\right) - 3 \left(x + 2\right)$

$\textcolor{w h i t e}{{x}^{3} + 2 {x}^{2} - 3 x - 6} = \left({x}^{2} - 3\right) \left(x + 2\right)$

$\textcolor{w h i t e}{{x}^{3} + 2 {x}^{2} - 3 x - 6} = \left(x - \sqrt{3}\right) \left(x + \sqrt{3}\right) \left(x + 2\right)$

So the remaining zeros are $- 2$ and $\pm \sqrt{3}$