# How do you use the rational root theorem to find the roots of 3x^4 + 5x^3 + 3x^2 - 7x + 9 = 0?

Jul 22, 2015

There are no rational roots.

#### Explanation:

Rational roots would be the factors of $\frac{9}{3}$ that is 3. These can be 1,-1,3,-3. None of these values satisfy the given equation, hence it is concluded that there are no rational roots.

Jul 22, 2015

From the rational root theorem we find that the only possible rational roots are $\pm \frac{1}{3}$, $\pm 1$, $\pm 3$ and $\pm 9$.

None of these is a root, so the equation has no rational roots.

#### Explanation:

By the rational root theorem, any rational root of $3 {x}^{4} + 5 {x}^{3} + 3 {x}^{2} - 7 x + 9 = 0$ must be of the form $\frac{p}{q}$ where $p$ and $q$ are integers, $q \ne 0$, $p$ a divisor of the constant term $9$ and $q$ a divisor of the coefficient $3$ of the term $3 {x}^{4}$ of highest degree.

That means that the only possible rational roots are:

$\pm \frac{1}{3}$, $\pm 1$, $\pm 3$ and $\pm 9$

Let $f \left(x\right) = 3 {x}^{4} + 5 {x}^{3} + 3 {x}^{2} - 7 x + 9$

Then:

$f \left(\frac{1}{3}\right) = \frac{65}{9}$

$f \left(- \frac{1}{3}\right) = \frac{311}{27}$

$f \left(1\right) = 13$

$f \left(- 1\right) = 17$

$f \left(3\right) = 393$

$f \left(- 3\right) = 165$

$f \left(9\right) = 23517$

$f \left(- 9\right) = 16353$

So none of the possible rational roots is a root of $f \left(x\right) = 0$

So $f \left(x\right) = 0$ has no rational roots.

In fact it has no real roots at all:

graph{3x^4+5x^3+3x^2-7x+9 [-20, 20, -4, 16]}