# How do you find the number of possible positive real zeros and negative zeros then determine the rational zeros given f(x)=x^4+2x^3-9x^2-2x+8?

Dec 19, 2016

The zeros of $f \left(x\right)$ are:

$1 , - 1 , - 4 , 2$

#### Explanation:

Given:

$f \left(x\right) = {x}^{4} + 2 {x}^{3} - 9 {x}^{2} - 2 x + 8$

Note that the pattern of the signs of the coefficients is $+ + - - +$. By Descartes' Rule of Signs, since this has $2$ changes of sign, $f \left(x\right)$ has $2$ or $0$ positive Real zeros.

The pattern of the signs of $f \left(- x\right)$ is $+ - - + +$. Since this has $2$ changes of sign, $f \left(x\right)$ has $2$ or $0$ negative Real zeros.

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 $8$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

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

In addition, note that the sum of the coefficients of $f \left(x\right)$ is $0$. That is:

$1 + 2 - 9 - 2 - 8 = 0$

Hence $f \left(1\right) = 0$, $x = 1$ is a zero and $\left(x - 1\right)$ a factor:

${x}^{4} + 2 {x}^{3} - 9 {x}^{2} - 2 x + 8 = \left(x - 1\right) \left({x}^{3} + 3 {x}^{2} - 6 x - 8\right)$

Note that $x = - 1$ is a zero of the remaining cubic, since:

$- 1 + 3 + 6 - 8 = 0$

So $\left(x + 1\right)$ is a factor:

${x}^{3} + 3 {x}^{2} - 6 x - 8 = \left(x + 1\right) \left({x}^{2} + 2 x - 8\right)$

Finally, note that $4 - 2 = 2$ and $4 \cdot 2 = 8$. So we find:

${x}^{2} + 2 x - 8 = \left(x + 4\right) \left(x - 2\right)$

So the zeros of $f \left(x\right)$ are:

$1 , - 1 , - 4 , 2$