How do you find all zeros with multiplicities of f(x)=-2x^3+19x^2-49x+20?

Mar 19, 2017

The zeros of $f \left(x\right)$ are $\frac{1}{2}$, $4$ and $5$, all with multiplicity $1$.

Explanation:

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

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

That means that the only possible rational zeros are:

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

in addition, notice that the signs of the coefficients are in the pattern $- + - +$. So by Descartes' Rule of Signs, since there are $3$ changes of sign, this cubic has $1$ or $3$ positive real zeros. Also the signs of the coefficients of $f \left(- x\right)$ are in the pattern $+ + + +$. So with no changes of sign, this cubic has no negative real zeros.

So the only possible rational zeros are:

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

Trying each in turn we first find:

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

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

$\textcolor{w h i t e}{f \left(\frac{1}{2}\right)} = 0$

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

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

To factor the remaining quadratic note that $4 + 5 = 9$ and $4 \cdot 5 = 20$.

Hence:

$- {x}^{2} + 9 x - 20 = - \left(x - 4\right) \left(x - 5\right)$

So the remaining two zeros are $x = 4$ and $x = 5$.

All of the zeros have multiplicity $1$.