# How do you use the Rational Zeros theorem to make a list of all possible rational zeros, and use the Descarte's rule of signs to list the possible positive/negative zeros of f(x)=x^3+4x^2-11x+6?

Jan 2, 2018

$x = - 6 , x = 1$

#### Explanation:

The pattern of signs of $f \left(x\right)$ is $+ + - +$ and it has two sign changes. Descartes' rule of signs then tells us that this polynomial has either no or two positive roots.

We then look at the pattern of $f \left(- x\right)$, which is $- + + +$, containing one change, meaning we have one negative root.

The rational roots theorem says that we can find all possible rational roots of a polynomial by dividing $\pm$ the factors of the last term by $\pm$ the factors of the first. In this case, we get:
$\pm 1 , \pm 2 , \pm 3 , \pm 6$

I will start by looking at the negative values, since we know for a fact the polynomial has a negative root. Of these, we find that $f \left(- 6\right)$ is a root. This means we can factor like so (using polynomial long division):
$\left(x + 6\right) \left({x}^{2} - 2 x + 1\right)$

The quadratic we just factored out is in the form of the expansion of ${\left(a - b\right)}^{2}$, so we can factor it:
$\left(x + 6\right) {\left(x - 1\right)}^{2}$

This means that $x = 1$ is also a root of the polynomial. You might wonder why we haven't got another root since Descartes' rule told us we'd have either two or no roots, but Descartes' rule counts the root $x = 1$ twice because it has a mulitplicity of two (due to the power of two).