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

##### 1 Answer

#### Answer:

The zeros of

#### Explanation:

By the rational roots theorem, any *rational* zeros of

That means that the only possible *rational* zeros are:

#+-1/2, +-1, +-2, +-5/2, +-4, +-5, +-10, +-20#

in addition, notice that the signs of the coefficients are in the pattern

So the only possible *rational* zeros are:

#1/2, 1, 2, 4, 5, 10, 20#

Trying each in turn we first find:

#f(1/2) = -2(1/8)+19(1/4)-49(1/2)+20#

#color(white)(f(1/2)) = (-1+19-98+80)/4#

#color(white)(f(1/2)) = 0#

So

#-2x^3+19x^2-49x+20 = (2x-1)(-x^2+9x-20)#

To factor the remaining quadratic note that

Hence:

#-x^2+9x-20 = -(x-4)(x-5)#

So the remaining two zeros are

All of the zeros have multiplicity