# What are all the possible rational zeros for #f(x)=3x^3-7x^2+29x-9# and how do you find all zeros?

##### 1 Answer

#### Answer:

The "possible" rational zeros are:

#+-1/3, +-1, +-3, +-9#

The actual zeros are:

#1/3# and#1+-2sqrt(2)i#

#### Explanation:

#f(x) = 3x^3-7x^2+29x-9#

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

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

#+-1/3, +-1, +-3, +-9#

Note also that the signs of the coefficients of

So the only possible rational real zeros are:

#1/3, 1, 3, 9#

Trying each in turn we immediately find:

#f(1/3) = 3(1/27)-7(1/9)+29(1/3)-9 = (1-7+87-81)/9 = 0#

So

#3x^3-7x^2+29x-9 = (3x-1)(x^2-2x+9)#

We can factor the remaining quadratic by completing the square, but it does require some Complex coefficients:

#x^2-2x+9 = x^2-2x+1+8#

#color(white)(x^2-9x+9) = (x-1)^2-(2sqrt(2)i)^2#

#color(white)(x^2-9x+9) = ((x-1)-2sqrt(2)i)((x-1)+2sqrt(2)i)#

#color(white)(x^2-9x+9) = (x-1-2sqrt(2)i)(x-1+2sqrt(2)i)#

Hence the other two zeros are:

#x = 1+-2sqrt(2)i#