# How do you find the number of possible positive real zeros and negative zeros then determine the rational zeros given f(x)=x^3-2x^2-8x?

Aug 8, 2018

The zeros of $f \left(x\right)$ are $0$, $- 2$ and $4$

#### Explanation:

Given:

$f \left(x\right) = {x}^{3} - 2 {x}^{2} - 8 x$

Note that the signs of the coefficients of $f \left(x\right)$ are in the pattern $+ - -$. With one change of sign, Descartes' Rule of Signs tells us that $f \left(x\right)$ has exactly one positive real zero.

The signs of the coefficients of $f \left(- x\right)$ are in the pattern $- - +$. With one change of sign, Descartes' Rule of Signs tells us that $f \left(x\right)$ has exactly one negative real zero.

In order to usefully apply the rational zeros theorem we need a non-zero constant term, but $f \left(x\right)$ has none, or in other words it is zero. That's because $x = 0$ is a zero of $f \left(x\right)$. Let's separate $x$ out as a factor first:

${x}^{3} - 2 {x}^{2} - 8 x = x \left({x}^{2} - 2 x - 8\right)$

Now applying the rational zeros theorem to ${x}^{2} - 2 x - 8$, any rational zeros of this quadratic are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 8$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that its only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 4 , \pm 8$

Trying these, we find:

${\left(\textcolor{b l u e}{- 2}\right)}^{2} - 2 \left(\textcolor{b l u e}{- 2}\right) - 8 = 4 + 4 - 8 = 0$

${\left(\textcolor{b l u e}{4}\right)}^{2} - 2 \left(\textcolor{b l u e}{4}\right) - 8 = 16 - 8 - 8 = 0$

So the zeros of $f \left(x\right)$ are $0$, $- 2$ and $4$