# 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^3-7x^2+x-7?

Jan 2, 2018

$x = 7 , x = \pm i$

#### Explanation:

The sign pattern of $f \left(x\right)$ is $+ - + -$. This is three changes, so Descartes' rule of signs tells us that we have either $3$ or $1$ positive root(s) to the polynomial.

The sign pattern of $f \left(- x\right)$ is $- - - -$, which has no changes, and therefor there are no negative roots to the polynomial.

The rational roots theorem tells us that we can find all rational roots to a polynomial by dividing $\pm$ the factors of the last term with $\pm$ the factors of the first term. In this case, we get:
$\pm 1 , \pm 7$

We also don't need to bother checking the negative roots, since Descartes' rule of signs told us there are no negative roots.

After checking $1$ and $7$, we see that only $f \left(7\right)$ is a solution. To find the remaining solutions (a cubic equation should have three solutions), we can use polynomial long division to factor:
$\left(x - 7\right) \left({x}^{2} + 1\right)$

Now we can solve for when the second factor equals $0$:
${x}^{2} + 1 = 0$

${x}^{2} = - 1$

$x = \pm \sqrt{- 1}$

$x = \pm i$

Now that we have all the zeroes we can see that our original conclusions with Descartes' rule were correct. We only got $1$ positive root, and no negative roots (Descartes' rule isn't concerned with complex roots).