# 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)=-2x^3+19x^2-49x+20?

Dec 22, 2017

See explanation...

#### Explanation:

Given:

$f \left(x\right) = - 2 {x}^{3} + 19 {x}^{2} - 49 x + 20$

By the rational zeros 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 $20$ and $q$ a divisor of the coefficient $- 2$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{2} , \pm 1 , \pm 2 , \pm \frac{5}{2} , \pm 4 , \pm 5 , \pm 10 , \pm 20$

The pattern of signs of the coefficients of $f \left(x\right)$ is $- + - +$. With $3$ changes, Descartes' Rule of Signs tells us that $f \left(x\right)$ has $3$ or $1$ positive real zeros.

The pattern of signs of the coefficients of $f \left(- x\right)$ is $+ + + +$. With no changes, Descartes' Rule of Signs tells us that $f \left(x\right)$ has no negative real zeros.

Putting these together, we can deduce that the only possible rational zeros of $f \left(x\right)$ are:

$\frac{1}{2} , 1 , 2 , \frac{5}{2} , 4 , 5 , 10 , 20$

Trying each in turn, we find:

$f \left(\frac{1}{2}\right) = - 2 \left(\textcolor{b l u e}{\frac{1}{8}}\right) + 19 \left(\textcolor{b l u e}{\frac{1}{4}}\right) - 49 \left(\textcolor{b l u e}{\frac{1}{2}}\right) + 20$

$\textcolor{w h i t e}{f \left(\frac{1}{2}\right)} = - \frac{1}{4} + \frac{19}{4} - \frac{98}{4} + \frac{80}{4} = 0$

So $x = \frac{1}{2}$ is a zero and $\left(2 x - 1\right)$ a factor:

$- 2 {x}^{3} + 19 {x}^{2} - 49 x + 20 = \left(2 x - 1\right) \left(- {x}^{2} + 9 x - 20\right)$

$\textcolor{w h i t e}{- 2 {x}^{3} + 19 {x}^{2} - 49 x + 20} = - \left(2 x - 1\right) \left(x - 4\right) \left(x - 5\right)$

So the three zeros are: $\frac{1}{2}$, $4$ and $5$