# What are all the possible rational zeros for #f(x)=32x^3-52x^2+17x+3# and how do you find all zeros?

##### 1 Answer

#### Explanation:

#f(x) = 32x^3-52x^2+17x+3#

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

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

#+-1/32, +-1/16, +-3/32, +-1/8, +-3/16, +-1/4, +-3/8, +-1/2, +-3/4, +-1, +-3/2, +-3#

That is rather a lot of possibilities to try, but note that the sum of the coefficients of

#32-52+17+3 = 0#

So

#32x^3-52x^2+17x+3 = (x-1)(32x^2-20x-3)#

We can find zeros of the remaining quadratic factor using an AC method:

Find a pair of factors of

The pair

#32x^2-20x-3#

#=32x^2-24x+4x-3#

#=(32x^2-24x)+(4x-3)#

#=8x(4x-3)+1(4x-3)#

#=(8x+1)(4x-3)#

Hence zeros: