# 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+2x^3-12x^2-40x-32?

Sep 23, 2017

Possible rational zeros:

$\pm 1 , \pm 2 , \pm 4 , \pm 8 , \pm 16 , \pm 32$

Descartes gives us that $f \left(x\right)$ has one positive and $3$ or $1$ negative zeros.

Actual zeros: $- 2 , - 2 , - 2 , 4$

#### Explanation:

Given:

$f \left(x\right) = {x}^{4} + 2 {x}^{3} - 12 {x}^{2} - 40 x - 32$

Rational roots theorem

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 $- 32$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means the the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 4 , \pm 8 , \pm 16 , \pm 32$

Descartes' Rule of Signs

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

The pattern of signs of coefficients of $f \left(- x\right)$ is $+ - - + -$. With $3$ changes of sign, Descartes' Rule of Signs allows us to deduce that $f \left(x\right)$ has $3$ or $1$ negative real zero.

Bonus - Find the actual zeros

Note that the coefficient of ${x}^{4}$ is odd, but all the other coefficients are even. Therefore any integer zero of $f \left(x\right)$ must be even.

We find:

$f \left(- 2\right) = {\left(- 2\right)}^{4} + 2 {\left(- 2\right)}^{3} - 12 {\left(- 2\right)}^{2} - 40 \left(- 2\right) - 32 = 16 - 16 - 48 + 80 - 32 = 0$

So $- 2$ is a zero and $\left(x + 2\right)$ a factor:

${x}^{4} + 2 {x}^{3} - 12 {x}^{2} - 40 x - 32 = \left(x + 2\right) \left({x}^{3} - 12 x - 16\right)$

We find that $- 2$ is also a zero of the remaining cubic expression:

${\left(- 2\right)}^{3} - 12 \left(- 2\right) - 16 = - 8 + 24 - 16 = 0$

So $\left(x + 2\right)$ is a factor again:

${x}^{3} - 12 x - 16 = \left(x + 2\right) \left({x}^{2} - 2 x - 8\right)$

Finally, to factor the remaining quadratic, note that $4 \cdot 2 = 8$ and $4 - 2 = 2$, so we find:

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

So:

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

has zeros $- 2$ with multiplicity $3$ and $4$ with multiplicity $1$.