# How do you use the rational roots theorem to find all possible zeros of P(x) = x^3 - 4x^2 + x + 6?

Jul 30, 2016

Zeros: $- 1 , 2 , 3$

#### Explanation:

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

By the rational roots theorem, any rational zeros of $P \left(x\right)$ are expressible as $\frac{p}{q}$ 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:

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

We find:

$P \left(- 1\right) = - 1 - 4 - 1 + 6 = 0$

So $x = - 1$ is a zero and $\left(x + 1\right)$ a factor:

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

We could simply evaluate $P \left(x\right)$ for the remaining possible zeros, but ${x}^{2} - 5 x + 6$ factors fairly easily:

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

So the other two zeros are $x = 2$ and $x = 3$.