# How do you find the rational roots of x^4+5x^3+7x^2-3x-10=0?

Jul 19, 2015

Use the rational roots theorem to find rational roots $x = 1$ and $x = - 2$.

#### Explanation:

Let $f \left(x\right) = {x}^{4} + 5 {x}^{3} + 7 {x}^{2} - 3 x - 10$

The rational roots theorem tells us that all rational roots of $f \left(x\right) = 0$ must be of the form $\frac{p}{q}$ where $p$ and $q$ are integers, $q \ne 0$, $p$ is a divisor of the constant term $10$ and $q$ is a divisor of the coefficient $1$ of the highest order term ${x}^{4}$.

So the possible rational roots are:

$\pm 1$, $\pm 2$, $\pm 5$ and $\pm 10$

Note that the sum of the coefficients of $f \left(x\right)$ is $0$:

$1 + 5 + 7 - 3 - 10 = 0$

So $f \left(1\right) = 0$.

That is $x = 1$ is a root of $f \left(x\right) = 0$ and $\left(x - 1\right)$ is a factor of $f \left(x\right)$

We also find

$f \left(- 2\right) = 16 - 40 + 28 + 6 - 10 = 0$

So $x = - 2$ is a root of $f \left(x\right) = 0$ and $\left(x + 2\right)$ is a factor of $f \left(x\right)$

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

We can long divide $f \left(x\right)$ by ${x}^{2} + x - 2$ to find:

$f \frac{x}{{x}^{2} + x - 2} = {x}^{2} + 4 x + 5$

A quick check of the discriminant of this quotient polynomial finds:

$\Delta = {4}^{2} - \left(4 \times 1 \times 5\right) = 16 - 20 = - 4$

So there are no more real roots and no more linear factors with real coefficients.