# What are all the possible rational zeros for f(x)=x^3+4x^2+5x+2 and how do you find all zeros?

##### 1 Answer
Sep 5, 2016

$f \left(x\right)$ has zeros $- 1$, $- 1$ and $- 2$.

#### Explanation:

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

By the rational roots theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $2$ and $q$ a divisor of the coefficient $1$ of the leading term.

So the only possible rational zeros are:

$\pm 1 , \pm 2$

Note also that $f \left(x\right)$ only has positive coefficients, so it has no positive Real zeros, which only leaves $- 1$ and $- 2$ as possible rational zeros.

We find:

$f \left(- 1\right) = - 1 + 4 - 5 + 2 = 0$

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

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

The only possible rational zeros of the remaining quadratic are also $- 1$ and $- 2$, with corresponding factors $\left(x + 1\right)$ and $\left(x + 2\right)$ and indeed:

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

So all the zeros of $f \left(x\right)$ are rational, being $- 1$ with multiplicity $2$ and $- 2$ with multiplicity $1$.