# How do you use the rational root theorem to list all possible roots for 2x^3+5x^2+4x+1=0?

Jan 15, 2017

The "possible" rational roots are: $\pm \frac{1}{2} , \pm 1$

The actual roots are: $- \frac{1}{2} , - 1 , - 1$

#### Explanation:

Given:

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

Note that the rational root theorem will only tell us about rational zeros, not necessarily all zeros.

By the rational root 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 $1$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{2}$, $\pm 1$

In this particular example note that all of the coefficients are positive. Hence $f \left(x\right)$ has no positive zeros, so let's check $f \left(- 1\right)$ first:

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

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

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

Note that $- 1$ is also a zero of $2 {x}^{2} + 3 x + 1$, since $2 - 3 + 1 = 0$:

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

So the third zero is $x = - \frac{1}{2}$

So in this particular example all of the roots were rational and therefore findable with the help of the rational roots theorem.