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

Aug 9, 2016

$P \left(x\right)$ has zeros $1 , - 2 , - 2$

#### Explanation:

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

By the rational root 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 $- 4$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 4$

We find:

$P \left(1\right) = 1 + 3 - 4 = 0$

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

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

The remaining quadratic is a perfect square trinomial:

${x}^{2} + 4 x + 4 = {\left(x + 2\right)}^{2}$

Hence the remaining zero is $- 2$ with multiplicity $2$.