Question

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

Answer 1

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

Explanation:

For `5x^4+9x^3+5x^2+2x+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, 1/5, 2/5, 4/5, -1, -2,-4, -1/5, -2/5, -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 `-1/5, -2/5, -4/5`, none of which are roots.

The equation has no rational roots.

Answer 2

Any rational roots must be `+-4,+-2,+-1,+-4/5,+-2/5` or `+-1/5`

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

Explanation:

Let `f(x) = 5x^4+9x^3+5x^2+2x+4`

By the rational root theorem, the rational roots of `f(x) = 0` must all be of the form `p/q` in lowest terms with `p` a divisor of `4` and `q` a divisor of `5`.

So the only possible rational roots are `+-4,+-2,+-1,+-4/5,+-2/5` or `+-1/5`.

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

In fact `f(x) = 0` has no real roots.

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