# How do you find the rational roots of x^4+3x^3+3x^2-3x-4=0?

Dec 9, 2015

Find rational roots $x = 1$ and $x = - 1$ by examining coefficients and separating out the corresponding factors.

#### Explanation:

First note that the sum of the coefficients is zero ($1 + 3 + 3 - 3 - 4 = 0$), so $x = 1$ is a root and $\left(x - 1\right)$ a factor:

${x}^{4} + 3 {x}^{3} + 3 {x}^{2} - 3 x - 4 = \left(x - 1\right) \left({x}^{3} + 4 {x}^{2} + 7 x + 4\right)$

The coefficients of the remaining factor are all positive, so it has no positive zeros. Notice that $- 1 + 4 - 7 + 4 = 0$, so $x = - 1$ is a root and $\left(x + 1\right)$ a factor:

${x}^{3} + 4 {x}^{2} + 7 x + 4 = \left(x + 1\right) \left({x}^{2} + 3 x + 4\right)$

The remaining quadratic factor is of the form $a {x}^{2} + b x + c$ with $a = 1$, $b = 3$ and $c = 4$. This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {3}^{2} - \left(4 \times 1 \times 4\right) = 9 - 16 = - 7$

Since this is negative, there are no more linear factors with Real coefficients and no more Real roots, let alone rational ones.

Alternatively, use the rational root theorem to find that the only possible rational roots are:

$\pm 1 , \pm 2 , \pm 4$

then just try each one in turn.