# How do you use the rational roots theorem to find all possible zeros of  P(x) = x^5 + 3x^3 + 2x - 6?

Jul 4, 2018

The "possible" rational zeros are $\pm 1 , \pm 2 , \pm 3 , \pm 6$.

The only real zero is $x = 1$

#### Explanation:

Given:

$P \left(x\right) = {x}^{5} + 3 {x}^{3} + 2 x - 6$

By the rational roots theorem any rational zeros of $P \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 6$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 3 , \pm 6$

In addition, note that the sum of the coefficients is zero, i.e.:

$1 + 3 + 2 - 6 = 0$

Hence we can deduce that $x = 1$ is a zero and $\left(x - 1\right)$ a factor:

${x}^{5} + 3 {x}^{3} + 2 x - 6 = \left(x - 1\right) \left({x}^{4} + {x}^{3} + 4 {x}^{2} + 4 x + 6\right)$

The remaining quartic has all positive coefficients, so any real zeros must be negative.

Trying $- 1$, $- 2$, $- 3$ and $- 6$ we find that none is a zero.

In fact it has only non-real complex zeros.