# How do you use the rational root theorem to find the roots of 5x^4 + 9x^3 + 5x^2 + 2x + 4 = 0?

Jul 4, 2015

List the possible rational roots and then test them. None of them are roots.

#### Explanation:

For $5 {x}^{4} + 9 {x}^{3} + 5 {x}^{2} + 2 x + 4 = 0$, we have leading coefficient $5$ and constant $4$.
The divisors of $5$ are $1$ and $5$.
The divisors of $4$ are $1$, $2$, and $4$.

So,the possible rational roots are:

$1 , 2 , 4 , \frac{1}{5} , \frac{2}{5} , \frac{4}{5} , - 1 , - 2 , - 4 , - \frac{1}{5} , - \frac{2}{5} , - \frac{4}{5}$

Since all of the coefficients are positive, there can be no positive roots.

Testing by synthetic division we can find that $- 1$ is not a root and for $- 2$, it is not a root and we get alternating signs, so $- 2$ is a lower bound on the roots,

So we need to test $- \frac{1}{5} , - \frac{2}{5} , - \frac{4}{5}$, none of which are roots.

The equation has no rational roots.

Jul 4, 2015

Any rational roots must be $\pm 4 , \pm 2 , \pm 1 , \pm \frac{4}{5} , \pm \frac{2}{5}$ or $\pm \frac{1}{5}$

None of these are roots, so it has no rational roots.

#### Explanation:

Let $f \left(x\right) = 5 {x}^{4} + 9 {x}^{3} + 5 {x}^{2} + 2 x + 4$

By the rational root theorem, the rational roots of $f \left(x\right) = 0$ must all be of the form $\frac{p}{q}$ in lowest terms with $p$ a divisor of $4$ and $q$ a divisor of $5$.

So the only possible rational roots are $\pm 4 , \pm 2 , \pm 1 , \pm \frac{4}{5} , \pm \frac{2}{5}$ or $\pm \frac{1}{5}$.

None of these are roots, so there are no rational roots.

In fact $f \left(x\right) = 0$ has no real roots.

graph{5x^4+9x^3+5x^2+2x+4 [-11.99, 11.42, -1.56, 10.15]}