# 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^4-9x^2-4x+12?

Apr 5, 2017

Zeros are $1$, $- 2$ (and $3$ and we have $f \left(x\right) = \left(x - 1\right) {\left(x + 2\right)}^{2} \left(x - 3\right)$

#### Explanation:

According to Rational zeros theorem, in a function such as

$f \left(x\right) = q {x}^{n} + {a}_{1} {x}^{\left(n - 1\right)} + \ldots + {a}_{\left(n - 1\right)} x + p$

zeros are factors of $\frac{p}{q}$.

Here we have $f \left(x\right) = {x}^{4} - 9 {x}^{2} - 4 x + 12$ i.e. $\frac{p}{q} = 12$ and hence zeros are $\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 6$ and $\pm 12$.

Note that as coefficients of polynomial add up to zero, $1$ is one zero.

Descarte's rule of signs is also method of determining the maximum number of positive and negative real roots of a polynomial.

Under this rule, for positive roots, we start with the sign of the coefficient of the lowest (or highest) power and count the number of sign changes, as you proceed from the lowest to the highest power (or highest to lowest). Here we should ignore powers that are not included i.e. whose coefficients are zeros.

If we have $n$ such changes, then we have $n$ or $n - 2$ or $n - 4$ such zeros.

Here in $f \left(x\right) = {x}^{4} \textcolor{red}{\hat{-}} 9 {x}^{2} - 4 x \textcolor{red}{\hat{+}} 12$, sign has changed twice. We have already seen that $1$ is a zero, so we must have another positive zero and trial gives us $3$ as other zero.

For negative zeros, we look at $f \left(- x\right)$ and as $f \left(- x\right) = {x}^{4} - 9 {x}^{2} + 4 x + 12$ and here sign changes takes place twice as $f \left(- x\right) = {x}^{4} \textcolor{red}{\hat{-}} 9 {x}^{2} \textcolor{red}{\hat{+}} 4 x + 12$. Hence, we have two or none negative zeros.

Trial tells us that $- 2$ is a zero, and hence we must have another negative number as zero. As we have zeros as $1$, $- 2$ and $3$, the fourth zero ought to be $- 2$ (as their product has to be $12$).

Hence, zeros are $1$, $- 2$ (twice) and $3$ and we have

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